19. Properties of Graphs

d. Second Derivative Information

2. Maxima & Minima and the Second Derivative Test

We previously looked at how to identify local mimima and local maxima using the First Derivative Test. While the First Derivative Test will always identify the local mimima and maxima, it is frequently easier to to use the Second Derivative Test:

Assume \(f(x)\) is twice differentiable on the interval \((a,b)\). Let \(c\) be a point in \((a,b)\) where \(f'(c)=0\).
If \(f''(c) \gt 0\), then \((c,f(c))\) is a local minimum.
If \(f''(c) \lt 0\), then \((c,f(c))\) is a local maximum.

thm_2nd_test

The condition \(f'(c)=0\) says the tangent line is horizontal at \(x=c\).
The condition \(f''(c) \gt 0\) says the curve is bending up at \(x=c\), and so stays above the tangent line. So \(x=c\) is the location of a local minimum.
The condition \(f''(c) \lt 0\) says the curve is bending down at \(x=c\), and so stays below the tangent line. So \(x=c\) is the location of a local maximum.

So to find the minima and maxima of \(f(x)\), we find the critical points \(x=c\) where \(f'(c)=0\) and compute the value of \(f''(c)\). If \(f''(c) \gt 0\), the curve is concave up at \(c\) and so \(c\) is a local minimum. If \(f''(c) \lt 0\), the curve is concave down at \(c\) and so \(c\) is a local maximum.

Find the local minima and local maxima of the function \(f(x)=\dfrac{x^4-39x^2-108}{3x}\) but using the Second Derivative Test.

In the example on the previous three pages we found the first and second derivatives are: \[\begin{aligned} f'(x)&=\dfrac{(x^2-4)(x^2-9)}{x^4} \\ f''(x)&=\dfrac{2(x^2-6)(x^2+6)}{x^3} \end{aligned}\] So the derivative is \(0\) at \(x=\pm2\) or \(x=\pm3\) and is undefined at \(x=0\) which are the critical points. However, \(x=0\) is also a point at which the function is undefined; so it cannot be a minimum or maximum. We evaluate the second derivative at each of the other critical points: \[\begin{aligned} f''(-3)&=\dfrac{2(9-6)(9+6)}{-27}=-\,\dfrac{10}{3} \lt 0& f''(3)&=\dfrac{2(9-6)(9+6)}{27}=\dfrac{10}{3} \gt 0 \\ f''(-2)&=\dfrac{2(4-6)(4+6)}{-8}=5 \gt 0& f''(2)&=\dfrac{2(4-6)(4+6)}{8}=-5 \lt 0 \end{aligned}\] So there are local maxima at \(x=-3\) and \(x=2\) and local minima at \(x=-2\) and \(x=3\) as shown on the numberline: eg_x^4-39x^2-108_3x_number_line_2deriv_maxmin

These are confirmed in the plot.

eg_x^4-39x^2-108_3x

If \(f''(c)=0\), the test FAILS! Further, we cannot say the point is a horizontal inflection point! It may still be a local minimum or maximum! The test just fails. We need to fall back on the First Derivative Test to distinguish between local minimum, local maximum or horizontal inflection point. This happens in the next exercise.

But first, here are the plots of \(4\) functions,
all of which have \(f'(2)=0\) — so \(x=2\) is a critical point, and
all of which have \(f''(2)=0\) — so the Second Derivative Test fails,
but one is a minimum, one is a maximum and two are horizontal inflection points:

rem_SDTfails_min
\(\begin{aligned} f(x)&=2+(x-2)^4 \\ f'(x)&=4(x-2)^3 \\ f'(2)&=0 \\ f''(x)&=12(x-2)^2 \\ f''(2)&=0 \end{aligned}\)
rem_SDTfails_max
\(\begin{aligned} f(x)&=2-(x-2)^4 \\ f'(x)&=-4(x-2)^3 \\ f'(2)&=0 \\ f''(x)&=-12(x-2)^2 \\ f''(2)&=0 \end{aligned}\)
rem_SDTfails_infl_incr
\(\begin{aligned} f(x)&=2+(x-2)^3 \\ f'(x)&=3(x-2)^2 \\ f'(2)&=0 \\ f''(x)&=6(x-2) \\ f''(2)&=0 \end{aligned}\)
rem_SDTfails_infl_decr
\(\begin{aligned} f(x)&=2-(x-2)^3 \\ f'(x)&=-3(x-2)^2 \\ f'(2)&=0 \\ f''(x)&=-6(x-2) \\ f''(2)&=0 \end{aligned}\)

Find the local minima, local maxima and horizontal inflection points of the function \(f(x)=\dfrac{1}{4}x^4-\dfrac{4}{3}x^3+2x^2\) but using the Second Derivative Test.

\(f(x)\) has a local minimum at \(x=0\) and a horizontal inflection point at \(x=2\). ex_x^4_4-4x^3_3+2x^2_number_line_2deriv_maxmininfl

In exercises on three previous pages, we found the first and second derivatives are: \[\begin{aligned} f'(x)&=x(x-2)^2 \\ f''(x)&=(3x-2)(x-2) \end{aligned}\] So the critical points are \(x=0\) and \(x=2\) and the secondary critical points are \(x=\dfrac{2}{3}\) and \(x=2\). To apply the Second Derivative Test, we evaluate the second derivative at each critical point: \[ f''(0)=(0-2)(0-2)=4 \qquad f''(2)=\left(3\cdot2-2\right)(2-2)=0 \] Since \(f''(0) \gt 0\), the function is concave up at \(x=0\) and so is a local minimum. Since \(f''(2)=0\), the Second Derivative Test fails and we turn to the First Derivative Test. Since \(f'(x)=x(x-2)^2\) is positive for all \(x \gt 0\), the function is increasing on both sides of \(x=2\) which is therefore a horizontal inflection point. The local minimum and the inflection point are indicated on the number line: ex_x^4_4-4x^3_3+2x^2_number_line_2deriv_maxmininfl

In the plot, we see the function has a minimum at \((0,0)\) where it is concave up and a horizontal inflection point at \((2,\dfrac{4}{3})\) where the function is increasing on both sides.

ex_x^4_4-4x^3_3+2x^2

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