Q. 5 - Calculus

Question

Can a point $x=c$ be both a local extremum and a critical point of a differentiable function f ? Both an inflection point and a critical point? Both an inflection point and a local extremum? Sketch examples, or explain why such a point cannot exist.

Step-by-Step Solution

Answer

The point $x=c$ satisfies the first two cases but does not satisfy the third case.

Step 1. Given Information.

The function $f$ is differentiable.

Step 2. A local extremum and a critical point.

Setting $f\text{'}\left(x\right)=0$for the critical points.

Using the first derivative we determine which critical points are local maxima, which are minima, and which are neither.

Using the graph we verify the statement.

At $x=1$they are both a local extremum point and a critical point.

Step 3. Inflection point and a critical point.

The function $f\left(x\right)=x^3$verifies the statement

Here, $x=0$ is both an inflection point and a critical point.

Step 4. Inflection point and a local extremum.

For the function $f\left(x\right)=x^3$, the critical point is $x=0$.

Using the points $x=-1,x=1$ for the numbers on either side of the critical point.

Now,

$f\text{'}\left(x\right)=3x^2$

Inputting $x=1$,

$$\begin{gathered} f\text{'}\left(1\right)=3{\left(1\right)}^2 \\ \Rightarrow f\text{'}\left(1\right)=3\ne 0 \end{gathered}$$

Inputting $x=-1$

$$\begin{gathered} f\text{'}(-1)=3{(-1)}^2 \\ \Rightarrow f\text{'}(-1)=3\ne 0 \end{gathered}$$

Therefore, $x=c$ cannot be both an inflection point and a local extremum.