Chapter 4, Problem 19RQ

If f′(x) exists and is nonzero for all x, then f(1) ≠ f(0).

To determine

Whether the statement, if $f^{\prime }\left(x\right)$ exists and non zero for all values of x then $f\left(1\right)\ne f\left(0\right)$ is true or false.

Answer to Problem 19RQ

The given statement is true.

Explanation of Solution

As $f^{\prime }\left(x\right)\ne 0$ there exist two cases.

Case 1:

Let $f^{\prime }\left(x\right)>0$ for all values of x .

Then the function f is strictly increasing function.

Therefore, $f\left(1\right)\ne f\left(0\right)$.

Case 2:

Let $f^{\prime }\left(x\right)<0$ for all values of x .

Then the function f is strictly decreasing function.

So, $f\left(1\right)\ne f\left(0\right)$.

Therefore, in both the cases $f\left(1\right)\ne f\left(0\right)$

Hence the given statement is true.