Question

State the converse of Theorem 3.10(a). Is the converse true? If so, explain why; if not, provide a counterexample.

Short answer

The converse of Theorem state that if f and f' are both differentiable on an interval Iand f is concave up on I then f'' is positive on I .

so converse theorem of $3.10\left(a\right)$is false because the second derivative might be zero at some point.

Take an example $f\left(x\right)=2x^3$where $f\text{'}\text{'}\left(x\right)=12x$and values of the second derivative is zero at zero and values at different points are following.

$$\begin{gathered} f\left(0\right)=0 \\ andf\text{'}\text{'}\left(1\right)=12 \\ f\text{'}\text{'}(-1)=-12 \end{gathered}$$

Step-by-step solution

Step 1. Given information.

The given theorem is as follows.

Supposef and f' are both differentiable on an interval I.

If f'' is positive on I, then f is concave up on I.

Step 2. Statement of the theorem.

The converse of Theorem state that if f and f' are both differentiable on an interval Iand f is concave up on I then f'' is positive on I .

Step 3. An explanation for converse.

If the second derivative f'' of a function f is Positive then f' is increasing on the interval, and if f' is increasing then f will concave up on interval.

Therefore f is concave up or down can be checked by the sign of its second derivative.

so converse theorem of $3.10\left(a\right)$is false because the second derivative might be zero at some point.

Consider a function $f\left(x\right)=2x^3.$

first derivative is $f\text{'}\left(x\right)=6x^2.$

the second derivative is $f\text{'}\text{'}\left(x\right)=12x.$

take several values of x.

$$\begin{gathered} f\text{'}\text{'}\left(0\right)=0 \\ andf\text{'}\text{'}\left(1\right)=12 \\ f\text{'}\text{'}(-1)=-12 \end{gathered}$$