Second Derivative Test

The second derivative test is used to determine if a given stationary point is a maximum or minimum.

The first step of the second derivative test is to find stationary points.

A stationary point is a point on a curve with a gradient of zero such that $f'(x) = 0$ or $\frac{dy}{dx} = 0$. $_\square$

Find the stationary points of the curve $y = \frac{1}{3}x^3 + 4x^2 - 20x$.

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We have $\frac{dy}{dx} = x^2 + 8x - 20.$ Since stationary points occur when $\frac{dy}{dx} = 0$, the values of $x$ for which $\frac{dy}{dx} = 0$ are obtained as follows:

$$\begin{aligned} x^2 + 8x - 20 &= 0\\ (x + 10)(x - 2) &= 0\\ x &= -10,~ x = 2. \end{aligned}$$

Substituting these values of $x$ back into the original curve equation gives the full pairs of coordinates:

$$\begin{aligned} y &= \frac{1}{3}(-10)^3 + 4(-10)^2 - 20(-10)\\& = -\frac{1000}{3} + 400 + 200 \\&= \frac{800}{3}\quad \\\\ y&= \frac{1}{3}(2)^3 + 4(2)^2 - 20(2) \\&= \frac{8}{3} + 16 - 40 \\&= -\frac{64}{3}, \end{aligned}$$

which implies that the stationary points are $\left(-10, \frac{800}{3}\right)$ and $\left(2, -\frac{64}{3}\right).$ $_\square$

Note in the example above that the full coordinates were found. When dealing with the second derivative test, only the $x$-coordinate values are needed.

A second derivative is found by differentiating a derivative of a function. In other words, differentiate a function to get its derivative, then differentiate again to get its second derivative.

Notation for second derivatives can be represented as

• $f''(x)$, read as "f double-dashed x";
• $\frac{d^2y}{dx^2}$, read as "d two y by d x squared."

Find the second derivative of $f(x) = 3x^4 + x^3 + 2x^2 + 17$.

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Differentiate the first time to get the derivative: $f'(x) = 12x^3 + 3x^2 + 4x.$

Now, differentiate the derivative to get the second derivative: $f''(x) = 36x^2 + 6x + 4$. $_\square$

Given that $p = -4q^5 - 3q^3 + q^2$, find $\frac{d^2p}{dq^2}$.

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First, find the derivative of $p:$ $\frac{dp}{dq} = -20q^4 - 9q^2 + 2q$.

Now, differentiate $\frac{dp}{dq}$ to find the second derivative: $\frac{d^2p}{dq^2} = -80q^3 - 18q + 2$. $_\square$

After finding the $x$-values of the stationary points, the second derivative needs to be found. Then, we substitute the $x$-values from the stationary points into the second derivative to perform the test. The test has three outcomes:

• If the second derivative is less than zero, the stationary point is a maximum.
• If the second derivative is greater than zero, the stationary point is a minimum.
• If the second derivative equals zero, the stationary point could be a point of inflection.

The derivative of a curve is found to be $g'(x) = 4x^2 - \frac{7}{2}x$. Stationary points on this curve occur when $x = 0$ and $x = \frac{7}{8}$. Determine if these points are maxima or minima using the second derivative test.

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We have $g''(x) = 8x - \frac{7}{2}$.
Now, substitute $x = 0$ to determine if it is a maximum or minimum point: $g''(0) = 8(0) - \frac{7}{2} = -\frac{7}{2}$.
As $g''(0) < 0$, the stationary point when $x = 0$ is a maximum.

Do the same for the other value: $g''\left(\frac{7}{8}\right) = 8\left(\frac{7}{8}\right) - \frac{7}{2} = 7 - \frac{7}{2} = \frac{7}{2}$.
As $g''\left(\frac{7}{8}\right) > 0$, the stationary point when $x = \frac{7}{8}$ is a minimum. $_\square$

• Differentiation Rules
• Implicit Differentiation
• Higher-order Derivatives