Question

Suppose that a function f is differentiate on [0, 1] and that its derivative is never zero. Using
mean value theorem, show that f(0) ≠f(1).​

Answer 1

Proof:

If a function f is differentiable in [0,1]

Then the function f will be continuous in [0,1]

By Mean value theorem we know that

If a function f(x) is continuous in [a,b] and differentiable in (a,b) then There exists a point c in the interval (a, b) such that:

$\boxed{f'(c)=\frac{f(b)-f(a)}{b-a}}$

Therefore,

For the given function

There exists a point c such that

$f'(c)=\frac{f(1)-f(0)}{1-0}$

or, $f'(c)=f(1)-f(0)$

Now if $f(1)=f(0)$

Then,

$f'(c)=0$

But it is given that the derivative of fuction f is never zero

Therefore, there is no c in [1, 0] such that

$f'(c)=0$

Therefore,

$f(1)\ne f(0)$

Hope this answer is helpful.

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Answer 2

F(1) is not equal to F(0)