Short and Sweet Calculus

## 4.5 Second derivative test for local extrema

Suppose $$c$$ is a critical number of a function $$f$$. To determine if $$f$$ has a local maximum, local minimum or neither at $$c$$, according to the First Derivative Test (Theorem 4.3) we need to find the sign of $$f’$$ near $$c$$. When determining the sign of $$f’$$ is difficult, we can use another test for local maximum and minimum values. This test is based on the geometrical observation that when the function has a horizontal tangent at $$c$$, if the function is concave down, the function has a local maximum at $$c$$, and if it is concave up, it has a local minimum (see Figure 4.14)

4.6. (Second Derivative Test) Suppose $$f'(c)=0$$.

(a) If $$f”(c)>0$$, then $$f$$ has a local minimum at $$c$$.

(b) If $$f”(c)<0$$, then $$f$$ has a local maximum at $$c$$.

• There are three situations where the Second Derivative Test is inconclusive:

1. $$f'(c)=f”(c)=0$$

2. $$f'(c)=0$$ and $$f”(c)$$ does not exist.

3. $$f'(c)$$ does not exist.

In these cases, $$c$$ may be a local minimum point, a local maximum point, or neither as shown (Figure 4.15) by the functions $f(x)=x^{4},\quad f(x)=-x^{4},\quad f(x)=x^{3}.$ For these functions, $$f'(0)=f”(0)=0$$, but $$x=0$$ is a point of local minimum for $$f(x)=x^{4}$$, a point of local maximum for $$f(x)=-x^{4}$$ and neither a local minimum nor maximum point for $$f(x)=x^{3}$$.

• Whenever the Second Derivative Test is inconclusive (as in the three situations discussed above) or when the second derivative is tedious to find, use the First Derivative Test to find the local extrema.

Example 4.6. Use the Second Derivative Test to find the local extrema of $f(x)=\frac{1}{2}x^{3}-\frac{3}{2}x^{2}+1.$

Solution

Let’s calculate the derivative to find the critical points of $$f$$: \begin{aligned} f'(x) & =\frac{3}{2}x^{2}-3x\\ & =\frac{3}{2}x(x-2)\end{aligned} Because $$f'(0)=f'(2)=0$$, $$x=0,2$$ are the critical points of $$f$$.

We previously solved this example using the First Derivative Test. Now we use the Second Derivative Test. $f'(x)=\frac{3}{2}x^{2}-3x\Rightarrow f”(x)=3x-3$ Because $f”(0)=3(0)-3=-3<0\quad(\Rightarrow f\text{ is concave down})$ it follows from the Second Derivative Test that $$f$$ has a local maximum at $$x=0$$. Similarly, $f”(2)=3(2)-3>0\quad(\Rightarrow f\text{ is concave up})$ and therefore, $$f$$ has a local minimum at $$x=2$$. The graph of $$f$$ is shown below.