Question

If y=f(x) is continuous on $[0,6]$, differentiable on $(0,6),f\left(0\right)=-2$ and $f\left(6\right)=16$, then at some point between $x=0$ and $x=6$, f'(x) must be equal to?

• A. $-18$
• B. $-3$
• C. $3$
• D. $14$

Answer 1

The correct option is C $3$

Here $y=f\left(x\right)$

now, at $x_1=0\Rightarrow y_1=f\left(x_1\right)$

$y_1=f\left(0\right)=-2$

$\therefore (0,-2)$ point is on function $f\left(x\right)$

and also at $x_2=6\Rightarrow y_2=f\left(x_2\right)$

$y_2=f\left(6\right)=16$

$\therefore (6,16)$ point is an function $f\left(x\right)$

So, we know that slope $\left(m\right)=\frac{dy}{dx}$

or $m=f\left(x\right)$

$\therefore \frac{y_2-y_1}{x_2-x_1}=m$

$\therefore f\left(x\right)=\frac{16-(-2)}{6-0}=\frac{18}{6}=3$

So , $f\left(x\right)=3$