Question

If anyone can help me??? Please!! How solve this?

Answer 1

We need to first check that
`f(x)` is continuous on the interval
`[0,3]`. If not, we're done. For this to happen, we need to have

`\displaystyle\lim_(x\to2^-)f(x)=\lim_(x\to2^+)f(x)=f(2)`

By the definition of
`f`,
`f(2)=2(2)-3=1`. The limits are then

`\displaystyle\lim_(x\to2^-)f(x)=\lim_(x\to2)2x-3=1`

`\displaystyle\lim_(x\to2^+)f(x)=\lim_(x\to2)6x-x^2-7=1`

The function is indeed continuous.

Next, we check for differentiability. For the derivative to exist everywhere, we need it to be continuous too. So we compute the derivatives of the pieces (note the strict inequalities):

`f'(x)=\begin{cases}2&\text{for }0<x<2\\6-2x&\text{for }2<x<3\end{cases}`

Judging by the first piece, in order for
`f'(x)` to be continuous on
`(0,3)`, the second piece must approach 2 as
`x\to2^+`, and if this happens, we can safely conclude that
`f'(2)=2`. We have

`\displaystyle\lim_(x\to2^+)f'(x)=\lim_(x\to2)6-2x=2`

and so
`f'(x)` is indeed continuous, and hence
`f(x)` is differentiable.

Finally, we need to find the value(s) of
`c` between 0 and 3 (exclusive) such that

`f'(c)=(f(3)-f(0))/(3-0)`

Since
`(f(3)-f(0))/(3-0)=\frac23`, and
`f'(x)=2` for
`0<x\le2`, we can expect
`f'(c)=\frac23` for some
`c` between 2 and 3:

`f'(c)=6-2c=\frac23\implies c=\frac83=2\frac23`