Question

If f is a continuously differentiable real-valued function defined on the open interval (-1,4) such that f(3)=5
and f'(x) = -1 for all x, what is the greatest possible value of f(0)​

Answer 1

Answer:

00000000 zero 0000000

Answer 2

Given,

f is a continuously differentiable real-valued function.

f(3)=5

and f'(x) = -1 for all x

To find,

The Greatest possible value of f(0)​.

Solution,

Since f′(x)≥−1, we have, for x≥0,

           $f(x)=f(0)+\int\limits^x_0 f'(t)dt\geq f(0)+\int\limits^x_0-1dt=f(0)-x$

from where we find

               f(0)≤x+f(x)

For x=3 we have

               f(0)≤8

To rule out everything else, identify a specific function f that meets all of the constraints and has the value f(0)≤8.

f(x)=-x+8 is an example of such a function.

Hence, 8 is the greatest possible value of f(0).