Derivative

1. Find \(f(x + h)\)
2. Find and simplify \(f(x + h) - f(x)\)
3. divide by \(h\) to get \(\frac{f(x + h) - f(x)} {h}\)
4. Let \(h\to 0\) : \(f'(x) = \lim\limits_{h \to 0} \frac {f(x + h) - f(x)} {h}\), if the limit exists.

• Example 1 : Constant Function

  \(f(x) = 10\). Find \(f'(x)\).

  \(f'(x) = \lim\limits_{h \to 0} \frac {f(x + h) - f(x)} {h}\)

  \(= \lim\limits_{h \to 0} \frac {10 - 10} {h}\)

  \(= \lim\limits_{h \to 0} 0\)

  \(= 0\)

• Example 2 : Polynomial Function

  \(f(x) = 3x^2 + 5x - 1\). Find \(f'(3)\).

  \(f'(x) = \lim\limits_{h \to 0} \frac {f(x + h) - f(x)} {h}\)

  \(= \lim\limits_{h \to 0} \frac {[3(x+h)^2 + 5(x+h) - 1] - (3x^2 + 5x - 1)} {h}\)

  \(= \lim\limits_{h \to 0} \frac {[3(x^2 + h^2 + 2xh) + 5(x+h) -1] - (3x^2 + 5x - 1)} {h}\)

  \(= \lim\limits_{h \to 0} \frac {3x^2 + 3h^2 + 6xh + 5x + 5h -1 - 3x^2 - 5x + 1} {h}\)

  \(= \lim\limits_{h \to 0} \frac {3h^2 + 6xh + 5h} {h}\)

  \(= \lim\limits_{h \to 0} 3h + 6x + 5\)

  \(= 6x + 5\)

  Note:

  To find \(f'(a)\), first find \(f'(x)\), and then substitute \(x\) by \(a\) in \(f'(x)\). The order can't be changed.

  • RIGHT

    \(f'(x) = 6x + 5\), so subsitute \(x\) by \(3\) and get \(f'(3) = 6\cdot 3 + 5 = 23\)

  • WRONG

    \(f(3) = 3\cdot 3^2 + 5\cdot 3 - 1 = 41\), which is a constant, so \(f'(3) = 0\).

• Example 3 : Sqrare Root Function : Using Conjugate

  \(f(x) = \sqrt{2x - 1} - 7\). Find \(f'(x)\).

  \(f'(x) = \lim\limits_{h \to 0} \frac {f(x + h) - f(x)} {h}\)

  \(= \lim\limits_{h \to 0} \frac {(\sqrt{2(x + h) - 1} - 7) - (\sqrt{2x - 1} - 7)} {h}\)

  \(= \lim\limits_{h \to 0} \frac {\sqrt{2(x + h) - 1} - \sqrt{2x - 1}} {h}\)

  \(= \lim\limits_{h \to 0} \frac {\sqrt{2(x + h) - 1} - \sqrt{2x - 1}} {h} \cdot \frac {\sqrt{2(x + h) - 1} + \sqrt{2x - 1}} {\sqrt{2(x + h) - 1} + \sqrt{2x - 1}}\)

  \(= \lim\limits_{h \to 0} \frac {[2(x + h) - 1] - (2x - 1)} {h} \cdot \frac {1} {\sqrt{2(x + h) - 1} + \sqrt{2x - 1}}\)

  \(= \lim\limits_{h \to 0} \frac {2x + 2h - 1 - 2x + 1} {h} \cdot \frac {1} {\sqrt{2(x + h) - 1} + \sqrt{2x - 1}}\)

  \(= \lim\limits_{h \to 0} \frac {2h} {h} \cdot \frac {1} {\sqrt{2(x + h) - 1} + \sqrt{2x - 1}}\)

  \(= \lim\limits_{h \to 0} \frac {2} {\sqrt{2(x + h) - 1} + \sqrt{2x - 1}}\)

  \(= \frac{2}{\sqrt{2x - 1} + \sqrt{2x - 1}}\)

  \(= \frac{2}{2\sqrt{2x - 1}}\)

  \(= \frac{1}{\sqrt{2x - 1}}\)

  Note:

  • Conjugate of \(\sqrt{2(x + h) - 1} - \sqrt{2x - 1}\) is \(\sqrt{2(x + h) - 1} + \sqrt{2x - 1}\).
  • \(\sqrt{2(x + h) - 1} - \sqrt{2x - 1} \neq \sqrt{2h}\)

• Example 4 : Fraction : Using Common Denominator

  \(f(x) = \frac{2}{3x+1}\). Find \(f'(x)\).

  \(f'(x) = \lim\limits_{h \to 0} \frac {f(x + h) - f(x)} {h}\)

  \(= \lim\limits_{h \to 0} \frac {\frac{2}{3(x+h)+1} - \frac{2}{3x+1}} {h}\) \(= \lim\limits_{h \to 0} \frac {\frac{2\cdot (3x+1)}{[3(x+h)+1] \cdot (3x+1)} - \frac{2\cdot[3(x+h)+1]}{[3(x+h)+1] \cdot (3x+1)}} {h}\) \(= \lim\limits_{h \to 0} \frac {\frac {2\cdot (3x+1) - 2\cdot[3(x+h)+1]} {[3(x+h)+1] \cdot (3x+1)}} {h}\)

  \(= \lim\limits_{h \to 0} \frac{2\cdot (3x+1) - 2\cdot[3(x+h)+1]} {h\cdot [3(x+h)+1] \cdot (3x+1)}\) \(= \lim\limits_{h \to 0} \frac{6x + 2 - 6x - 6h - 2} {h\cdot [3(x+h)+1] \cdot (3x+1)}\) \(= \lim\limits_{h \to 0} \frac{- 6h} {h\cdot [3(x+h)+1] \cdot (3x+1)}\)

  \(= \lim\limits_{h \to 0} \frac{- 6} {[3(x+h)+1] \cdot (3x+1)}\) \(= \frac{- 6} {(3x+1) \cdot (3x+1)}\) \(= \frac{- 6} {(3x+1)^2}\)

If \(f(x) = g(x) \pm h(x)\) and if \(g'(x)\) and \(h'(x)\) both exist, then \[ f'(x) = g'(x) \pm h'(x) \]

If \(f(x) = k g(x)\) where \(k\) is a real number and \(g'(x)\) exists, then \[ f'(x) = k g'(x) \]

If \(f(x) = g(x) \cdot h(x)\) and if \(g'(x)\) and \(h'(x)\) both exist, then \[ f'(x) = g'(x) \cdot h(x) + g(x) \cdot h'(x)\]

• Example

  \(f(x) = g(x) \cdot h(x)\) where \(g(x) = 2x^3 + 1\) and \(h(x) = 7x^2 - 2\)

  \(g'(x) = 2\cdot 3\cdot x^2 = 6x^2\)

  \(h'(x) = 7\cdot 2\cdot x = 14x\)

  \(f'(x) = g'(x) \cdot h(x) + g(x) \cdot h'(x)\)

  \(= 6x^2 (7x^2 - 2) + (2x^3 + 1) \cdot 14x\)

  \(= 42 x^4 - 12 x^2 + 28 x^4 + 14x\)

  \(= 70 x^4 - 12 x^2 + 14x\)

If \(f(x) = \frac {g(x)} {h(x)}\) and if \(g'(x)\) and \(h'(x)\) both exist, then \[f'(x) = \frac {g'(x) \cdot h(x) - g(x) \cdot h'(x)} {[h(x)]^2} \]

• Example

  \(f(x) = \frac{g(x)} {h(x)}\) where \(g(x) = 2x^3 + 1\) and \(h(x) = 7x^2 - 2\)

  \(g'(x) = 2\cdot 3\cdot x^2 = 6x^2\)

  \(h'(x) = 7\cdot 2\cdot x = 14x\)

  \(f'(x) = \frac {g'(x) \cdot h(x) - g(x) \cdot h'(x)} {[h(x)]^2}\)

  \(= \frac {6x^2 \cdot (7x^2 - 2) - (2x^3+1) \cdot 14x} {(7x^2-2)^2}\)

  \(= \frac {6x^2 (7x^2 - 2) - (2x^3 + 1) \cdot 14x} {(7x^2-2)^2}\)

  \(= \frac {42 x^4 - 12 x^2 - 28 x^4 - 14x} {(7x^2-2)^2}\)

  \(= \frac {14 x^4 - 12 x^2 - 14x} {(7x^2-2)^2}\)

For a composite function \(f(g(x))\), if we regard the function \(g(x)\) as a single variable \(g\) by letting \(g = g(x)\), then \[ \frac{df(g(x))}{dx} = \frac{df(g)}{dg} \cdot \frac{dg(x)}{dx} \]

• \(g\) and \(g(x)\)
  • when writting write \(g\), we mean a variable \(g\), just like \(x, u\) and \(v\)
  • when writing \(g(x)\), we mean a function of \(x\)
• \(\frac{df(g)}{dg}\) means
  • the variable of function \(f(g)\) is \(g\)
  • The derivative of \(f(g)\) is with respect to the variable \(g\)
• In short notation, we may memorize the chain rule by (plausibly, you get the LHS by cancelling \(dg\) on the RHS) \[ \frac{df}{dx} = \frac{df}{dg} \cdot \frac{dg}{dx} \]

Let \(f(x) = (2x^2 + 1)^3\). Find \(\frac{df(x)}{dx}\).

Clearly \(f(x)\) is a composite function, where

• The inner function is \(g(x) = 2x^2 + 1\).
• The outer function is \(f(g) = g^3\) where \(g=g(x)\).

By Chain Rule \(\frac{df(g(x))}{dx} = \frac{df(g)}{dg} \cdot \frac{dg(x)}{dx}\), to find \(\frac{df(g(x))}{dx}\), we need \(\frac{df(g)}{dg}\) and \(\frac{dg(x)}{dx}\).

• By regarding \(g\) as a single variable, we find \(\frac{df(g)}{dg} = \frac{d(g^3)}{dg} = 3 g^2\) by power rule
• By regarding \(x\) as the variable, we find \(\frac{dg(x)}{dx} = \frac{d(2x^2 + 1)}{dx} = 4x\) by power rule
• Putting them into the formula of chain rule, we get \(\frac{df(g(x))}{dx} = 3 g^2 \cdot 4x = 12xg^2\)
• Here \(g\) is a variable that is not given by the question, so we have to replace it. Since \(g = g(x)\), we just replace \(g\) with the expression of \(g(x)\), which is \(2x^2 + 1\) here. Thus the final answer is \(\frac{df(g(x))}{dx} = 12x(2x^2+1)^2\).

If \(f(x) = k\) where \(k\) is a real number, then \(f'(x) = 0\).

If \(f(x) = x^n\) where n is a real number, then \(f'(x) = n\cdot x^{n-1}\)

/Remark: / Whenever you meet something like \(\frac{1}{x}\) or \(\sqrt{x}\), rewrite them as \(x^{-1}\) or \(x^{\frac{1}{2}}\) respectively.

By the power rule and addition rule, we get that for a polynomial \(f(x) = a_0 + a_1 x + a_2 x^2 + \dots + a_p x^p\) where \(a_0, a_1, a_2, \dots, a_p\) are constants, the derivative is

\(f'(x) = (a_0 + a_1 x + a_2 x^2 + \dots + a_p x^p)'\)

\(= (a_0)' + (a_1 x)' + (a_2 x^2)' + \dots + (a_p x^p)'\)

\(= 0 + a_1 + a_2 \cdot 2x + a_3 \cdot 3x^2 + \dots + a_p \cdot px^{p-1}\)

/Example: / \((5x^3 + 7x^2 - 3x + 2)' = 5 \cdot 3 x^2 + 7 \cdot 2x - 3\).

For any constant \(a\) such that \(a>0\) and \(a\neq 1\), \[ \frac{d}{dx} a^x = lna \cdot a^x \] \[ \frac{d}{dx} e^x = e^x \]

• Examples
  • \(\frac{d}{dx} 3^x = ln3 \cdot 3^x\)

  • \(\frac{d}{dx} 3^{2x^3} =\) ?

    Let \(f(x) = 3^{2x^3}\), then we can see it is a composite function where

    • the inner function \(g(x) = 2x^3\), with \(\frac{dg(x)}{dx} = 6x^2\)
    • the outer function \(f(g) = 3^g\), with \(\frac{df(g)}{dg} = ln3 \cdot 3^g\) where \(g = g(x)\)

    so we should apply the Chain Rule and get

    \(\frac{df(g(x))}{dx} = \frac{df(g)}{dg} \cdot \frac{dg(x)}{dx} = ln3 \cdot 3^g \cdot 6x^2 = 6 \cdot ln3 \cdot x^2 \cdot 3^{2x^3}\)

  • \(\frac{d}{dx} e^{4x^3 + 2} =\) ?

    Let \(f(x) = e^{4x^3 + 2}\), then we can see it is a composite function where

    • the inner function \(g(x) = 4x^3 + 2\), with \(\frac{dg(x)}{dx} = 4 \cdot 3x^2 + 0 = 12 x^2\)
    • the outer function \(f(g) = e^g\), with \(\frac{df(g)}{dg} = e^g\) where \(g = g(x)\)

    so we should apply the Chain Rule and get

    \(\frac{df(g(x))}{dx} = \frac{df(g)}{dg} \cdot \frac{dg(x)}{dx} = e^g \cdot 12x^2 = e^{4x^3 + 2} 12x^2\)

  • \(\frac{d}{dx}e^2=\) ?

    Caution : \(e^2\) is a constant, so \(\frac{de^2}{dx} = e^2\) is WRONG. The exponential rule does not apply because \(2\) is not the variable \(x\). Instead, we should use the constant rule to get \(\frac{de^2}{dx} = 0\).

For any constant \(a\) such that \(a>0\) and \(a\neq 1\), \[ \frac{d}{dx} log_ax = \frac{1}{x \cdot lna} \] \[ \frac{d}{dx} lnx = \frac{1}{x} \]

• Examples
  • \(\frac{d}{dx} log_2 x = \frac{1}{x \cdot ln2}\)

  • \(\frac{d}{dx} log_2 x^3 =\) ?

    Let \(f(x) = log_2 x^3\), then we can see it is a composite function where

    • the inner function \(g(x) = x^3\), with \(\frac{dg(x)}{dx} = 3x^2\)
    • the outer function \(f(g) = log_2 g\), with \(\frac{f(g)}{dg} = \frac{1}{g \cdot ln2}\)

    so we should apply the Chain Rule and get

    \(\frac{df(g(x))}{dx} = \frac{df(g)}{dg} \cdot \frac{dg(x)}{dx} = \frac{1}{g \cdot ln2} \cdot 3x^2 = \frac{3x^2}{x^3 \cdot ln2}\)

  • \(\frac{d}{dx} ln(3x^2 - 1)=\) ?

    Let \(f(x) = ln (3x^2 - 1)\), then we can see it is a composite function where

    • the inner function \(g(x) = 3x^2 - 1\), with \(\frac{dg(x)}{dx} = 3\cdot 2x - 0 = 6x\)
    • the outer function \(f(g) = ln(g)\), with \(\frac{f(g)}{dg} = \frac{1}{g}\)

    so we should apply the Chain Rule and get

    \(\frac{df(g(x))}{dx} = \frac{df(g)}{dg} \cdot \frac{dg(x)}{dx} = \frac{1}{g} \cdot 6x = \frac{6x}{3x^2 - 1}\)

  • \(\frac{d}{dx}ln2=\) ?

    Caution : \(ln2\) is a constant, so \(\frac{dln2}{dx} = \frac{1}{2}\) is WRONG. The logarithmic rule does not apply because \(2\) is not the variable \(x\). Instead, we should use the constant rule to get \(\frac{dln2}{dx} = 0\).

• Remark

  Comparison between Power rule and Exponential Rule

  • Power Rule : \((x^a)' = a\cdot x^{a-1}\) requires variable \(x\) be the base and constant \(a\) be the power
  • Exponential Rule : \(a^x = lna \cdot a^x\) requires constant \(a\) \((a>0\) and \(a\neq 1)\) be the base and variable \(x\) be the power

For example, if \(f(x) = ln \left( \frac{e^{3x^2 + 1} \cdot (2x+5)^2 \cdot \sqrt{3x-1}}{(4x+1)^3} \right)\), then we may use the logarithmic property to rewrite it as

\(f(x) = ln \left( \frac{e^{3x^2 + 1} \cdot (2x+5)^2 \cdot \sqrt{3x-1}}{(4x+1)^3} \right)\)

\(= ln \left( \frac{e^{3x^2 + 1} \cdot (2x+5)^2 \cdot \sqrt{3x-1}}{(4x+1)^3} \right)\)

\(= ln(e^{3x^2 + 1}(2x+5)^2 \cdot \sqrt{3x-1}) - ln((4x+1)^3)\)

\(= ln(e^{3x^2 + 1}) + ln((2x+5)^2) + ln(\sqrt{3x-1}) - ln((4x+1)^3)\)

\(= (3x^2 + 1) lne + 2ln(2x+5) + \frac{1}{2} ln(3x-1) - 3ln(4x+1)\)

\(= 3x^2 + 1 + 2ln(2x+5) + \frac{1}{2} ln(3x-1) - 3ln(4x+1)\)

Now it is much easier for us to find the derivative.

Using relevant rules, we get

\(\frac{d}{dx} f(x)\)

\(= \frac{d}{dx} [3x^2 + 1 + 2ln(2x+5) + \frac{1}{2} ln(3x-1) - 3ln(4x+1)]\)

\(= 3 \frac{d}{dx} x^2 + \frac{d1}{dx} + 2 \frac{d}{dx} ln(2x+5) + \frac{1}{2} \frac{d}{dx} ln(3x-1) - 3 \frac{d}{dx} ln(4x+1)\)

\(= 3\cdot 2x + 0 + 2 \cdot \frac{2}{2x+5} + \frac{1}{2} \cdot \frac{3}{3x-1} - 3 \cdot \frac{4}{4x+1}\)

\(= 6x + \frac{4}{2x+5} + \frac{3}{2(3x-1)} - \frac{12}{4x+1}\)

For example, it is not a good idea to directly use quotient rule when we find the derivative of \(f(x) = \frac{x^3 + 2x^2 + 3x + 4}{5x^2}\). Instead, we may split the long numerator and then simplify each term:

\(f(x) = \frac{x^3 + 2x^2 + 3x + 4}{5x^2}\)

\(= \frac{x^3}{5x^2} + \frac{2x^2}{5x^2} + \frac{3x}{5x^2} + \frac{4}{5x^2}\)

\(= \frac{x}{5} + \frac{2}{5} + \frac{3}{5}x^{-1} + \frac{4}{5}x^{-2}\)

In this way, we immediately get the the derivative of \(f(x)\) as

\(f'(x)\)

\(= \left( \frac{x}{5} \right)' + \left( \frac{2}{5} \right)' + \left( \frac{3}{5} x^{-1} \right)' + \left( \frac{4}{5} x^{-2} \right)'\)

\(= \frac{1}{5} + 0 + \frac{3}{5} \cdot (-1) x^{-2} + \frac{4}{5} \cdot (-2) x^{-3}\)

\(= \frac{1}{5} - \frac{3}{5} x^{-2} - \frac{8}{5} x^{-3}\)

The critical numbers for a function \(f\) are those numbers \(c\) in the domain of \(f\) for which

• \(f'(c) = 0\) or
• \(f'(c)\) does not exist

A critical point is a point whose \(x\)-coordinate is the critical number \(c\) and whose \(y\)-coordinate is \(f(c)\).

For \(f(x) = \frac{x^2 + 1}{2x - 1}\), by quotient rule we have

\(f'(x)\)

\(= \left( \frac{x^2 + 1}{2x - 1} \right)'\)

\(= \frac{(x^2 + 1)'(2x - 1) - (x^2 + 1)(2x - 1)'}{(2x - 1)^2}\)

\(= \frac{2x(2x - 1) - (x^2 + 1) \cdot 2}{(2x - 1)^2}\)

\(= \frac{4x^2 - 2x - 2x^2 - 2}{(2x - 1)^2}\)

\(= \frac{2x^2 - 2x - 2}{(2x - 1)^2}\)

• \(f'(x)\) is undefined at \(x = \frac{1}{2}\), but \(\frac{1}{2}\) is NOT a critical number of \(f(x)\) because it is NOT in the domain

• Letting \(f'(x) = 0\) means letting \(2x^2 - 2x - 2 = 0\). By formula for quadratic roots (the roots of \(ax^2 + bx + c = 0\) are \(x = \frac{-b \pm \sqrt{4ac}}{2a}\)), we know

  \(\frac{-(-2) \pm \sqrt{(-2)^2 - 4 \cdot 2 \cdot (-2)}}{2\cdot 2} = \frac{2 \pm \sqrt{20}}{4} = \frac{2 \pm 2 \sqrt{5}}{4} = \frac{1 \pm \sqrt{5}}{2}\) are 2 critical numbers.

• Hence, only \(\frac{1 + \sqrt{5}}{2}\) and \(\frac{1 - \sqrt{5}}{2}\) are critical numbers of \(f(x)\), and \(\left( \frac{1 + \sqrt{5}}{2}, f(\frac{1 + \sqrt{5}}{2}) \right)\) and \(\left( \frac{1 - \sqrt{5}}{2}, f(\frac{1 - \sqrt{5}}{2}) \right)\) are critical points. (Calculate \(f(\frac{1 - \sqrt{5}}{2})\) and \(f(\frac{1 + \sqrt{5}}{2})\) by yourself. The same for all following examples.)

• Find all critical numbers of \(f(x)\)
• Separate the whole domain of \(f(x)\) into several intervals by the critical numbers
• Within each interval, choose a value of \(x\), say \(a\), and see whether \(f'(a) > 0, < 0\) or \(= 0\).
• Decide where \(f(x)\) is increasing or decreasing on each interval
  • If \(f'(a) > 0\), then \(f(x)\) is increasing on that interval
  • If \(f'(a) < 0\), then \(f(x)\) is decreasing on that interval
  • If \(f'(a) = 0\), then \(f(x)\) is constant on that interval

• We have already found that the \(f(x) = \frac{x^2 + 1}{2x - 1}\) is defined on \((-\infty, \frac{1}{2}) \cup (\frac{1}{2}, \infty)\) and critical numbers of \(f(x)\) are \(\frac{1 + \sqrt{5}}{2}\) and \(\frac{1 - \sqrt{5}}{2}\).
• These critical numbers separate the domain of \(f(x)\) into the 1st row of intervals in the following table. Now we have to fill in the 2nd and the 3rd rows.
• Within each interval, we choose a number \(a\) and evaluate \(f'(a)\) to see its sign

• Based on the sign of \(f'(a)\), we determine whether the function is increasing or decreasing on that interval

  \(x\)	\((-\infty\), \(\frac{1 - \sqrt{5}}{2})\)	\((\frac{1 - \sqrt{5}}{2}, \frac{1}{2})\)	\((\frac{1}{2}, \frac{1 + \sqrt{5}}{2})\)	\((\frac{1 + \sqrt{5}}{2}, \infty)\)
  \(f'(x)\)	\(f'(-10) > 0\)	\(f'(0) < 0\)	\(f'(1) < 0\)	\(f'(10) > 0\)
  \(f(x)\)	increasing	desreasing	decreasing	increasing

Let \(c\) be a critical number for function \(f\). Suppose that \(f\) is continous on \((a, b)\) and differentiable on \((a, b)\) except possibly at \(c\), and that \(c\) is the only critical number for \(f\) in \((a, b)\). Then

• \(f(c)\) is a relative maximum of \(f\) if \(f'(x) > 0\) in the interval \((a, c)\) and \(f'(x) < 0\) in the interval \((c, b)\)
• \(f(c)\) is a relative minimum of \(f\) if \(f'(x) < 0\) in the interval \((a, c)\) and \(f'(x) > 0\) in the interval \((c, b)\)

Suppose \(f''(x)\) exists on some open interval containing \(c\) except possibly at \(c\) itself, and suppose \(f'(c) = 0\).

• If \(f''(c) > 0\), then \(f(c)\) is a relative minimum
• If \(f''(c) < 0\), then \(f(c)\) is a relative maximum
• If \(f''(c) = 0\) or \(f''(c)\) does not exist, then the test gives no information about extrema, so use the first derivative test. (Remark : so, if possible, always use First Derivative Test.)

• Find all critical numbers of function \(f(x)\), and separate the domain into several intervals by these critical numbers
• At each critical number, use First Derivative Test or Second Derivative Test to determine whether it gives a relative maximum or minimum or no information.

• Using First Derivative Test

  For function \(f(x) = \frac{x^2 + 1}{2x - 1}\), we already know the critical numbers are \(\frac{1 + \sqrt{5}}{2}\) and \(\frac{1 - \sqrt{5}}{2}\). Also, we have already got the following table, so we can directly apply the First Derivative Test.

  \(x\)	\((-\infty\), \(\frac{1 - \sqrt{5}}{2})\)	\((\frac{1 - \sqrt{5}}{2}, \frac{1}{2})\)	\((\frac{1}{2}, \frac{1 + \sqrt{5}}{2})\)	\((\frac{1 + \sqrt{5}}{2}, \infty)\)
  \(f'(x)\)	\(f'(-10) > 0\)	\(f'(0) < 0\)	\(f'(1) < 0\)	\(f'(10) > 0\)
  \(f(x)\)	increasing	desreasing	decreasing	increasing

  • On \((-\infty\), \(\frac{1 - \sqrt{5}}{2})\), \(f'(x) > 0\) and \(f(x)\) is increasing. On \((\frac{1 - \sqrt{5}}{2}, \frac{1}{2})\), \(f'(x) < 0\) and \(f(x)\) is decreasing. Thus \(f(\frac{1 - \sqrt{5}}{2})\) is a relative maximum.
  • On \((\frac{1}{2}, \frac{1 + \sqrt{5}}{2})\), \(f'(x) < 0\) and \(f(x)\) is decreasing. On \((\frac{1 - \sqrt{5}}{2}, \infty)\), \(f'(x) > 0\) and \(f(x)\) is increasing. Thus \(f(\frac{1 + \sqrt{5}}{2})\) is a relative minimum.
  • \(f(\frac{1}{2})\) is ignored as it is undefined.

• Using Second Derivative Test

  For function \(f(x) = \frac{x^2 + 1}{2x - 1}\), we already know \(f'(x) = \frac{2x^2 - 2x - 2}{(2x - 1)^2}\), so the second derivative is

  \(f''(x) = [f'(x)]'\)

  \(= \left( \frac{2x^2 - 2x - 2}{(2x - 1)^2} \right)'\)

  \(= \frac{(2x^2 - 2x - 2)'(2x-1)^2 - (2x^2 - 2x - 2)[(2x-1)^2]'}{(2x - 1)^4}\)

  \(= \frac{(4x - 2)(2x-1)^2 - (2x^2 - 2x - 2)[2\cdot 2 \cdot (2x-1)]}{(2x - 1)^4}\)

  \(= \frac{(4x - 2)(2x-1)^2 - 4(2x^2 - 2x - 2) (2x-1)}{(2x - 1)^4}\)

  \(= \frac{(4x - 2)(2x-1) - 4(2x^2 - 2x - 2)}{(2x - 1)^3}\) [by cancelling \((2x-1)\)]

  \(= \frac{8x^2 - 4x - 4x + 2 - 8x^2 + 8x + 8}{(2x - 1)^3}\)

  \(= \frac{10}{(2x-1)^3}\)

  Thus we have

  • \(f'(\frac{1 - \sqrt{5}}{2}) = 0\) and \(f''(\frac{1 - \sqrt{5}}{2}) < 0\), so \(f(\frac{1 - \sqrt{5}}{2})\) is a relative maximum.
  • \(f'(\frac{1 + \sqrt{5}}{2}) = 0\) and \(f''(\frac{1 + \sqrt{5}}{2}) > 0\), so \(f(\frac{1 + \sqrt{5}}{2})\) is a relative minimum.

• Find all values of \(x\) such that \(f''(x) = 0\) or \(f''(x)\) does not exist (these values are called possible inflection points, see next section)
• Separate the whole domain of \(f(x)\) into several intervals by these numbers
• Within each interval, choose a value of \(x\), say \(a\), and see whether \(f''(a) > 0, < 0\) or \(= 0\).
• Decide where \(f(x)\) is concave upward or downward on each interval
  • If \(f''(a) > 0\), then \(f(x)\) is concave upward on that interval
  • If \(f''(a) < 0\), then \(f(x)\) is concave downward on that interval

For function \(f(x) = \frac{x^2 + 1}{2x - 1}\), we already know \(f''(x) = \frac{10}{(2x - 1)^3}\).

• Find all points where \(f''(x) = 0\) or \(f''(x)\) is undefined.
  • \(f''(x) = 0\) never holds because the numerator \(10\) can never equal \(0\).
  • \(f''(x)\) is undefined at \(x=\frac{1}{2}\)
  • Therefore, \(\frac{1}{2}\) is the only possible inflection point
• The domain of the original function \(f(x)\) is \((-\infty,\frac{1}{2}) \cup (\frac{1}{2},\infty)\), so the possible inflection point \(x=\frac{1}{2}\) doesn't separate the domain any more. As a result, the first row of the following table contains only two intervals.
• Calculate \(f''(x)\) at some special point within each interval.

• Based on the sign of the second row of the table, we can determine the concavity of the function \(f(x)\) as follows.

  \(x\)	\((-\infty, \frac{1}{2})\)	\((\frac{1}{2}, \infty)\)
  \(f''(x)\)	\(f''(0) < 0\)	\(f''(1) > 0\)
  \(f(x)\)	concave downward	concave upward

• Find all possible inflection points (candidates for further investigation). A point \(c\) is said to be a possible inflection point if \(f''(c) = 0\) or \(f''(c)\) is not defined.
• At each possible inflection point, determine by the above method whether it is truly an inflection point.

For function \(f(x) = \frac{x^2 + 1}{2x - 1}\), we already know \(f''(x) = \frac{10}{(2x - 1)^3}\). \(f''(x)\) is undefined at \(x=\frac{1}{2}\), and \(f''(x) = 0\) can not be achieved for any value of \(x\).

• Thus \(\frac{1}{2}\) is the only possible inflection points.
• At \(\frac{1}{2}\), the concavity changes, so \(\frac{1}{2}\) is an inflection point.