Question

The graph of y = f ′(x), the derivative of f(x), is shown below. Given f(4) = 6, evaluate f(0).

Answer 1

Answer:

B. 2

Explanation:

From the graph we get

(1). $f'(x)=x$ for  $-2\leq x\leq 2$

(2). $f'(x)=-x+4$ for  $x>2$

(3). $f'(x)=-x-4$ for $x$

Now let us take the anti derivative of the second function, and we get

$f(x)=-\frac{x^2}{2} +4x+c$ for $x>2$

We find $c$ from the condition $f(4)=6$:

$f(4)=-\frac{4^2}{2} +(4*4)+c=6$

$8+c=6$

$\boxed{c=-2}$

Thus we have

$f_2(x)=-\frac{x^2}{2} +4x-2$.

Now we find the anti derivative of the first function, and get

$f_1(x)=\frac{x^2}{2} +c_1$

What we note now is that the function is continuous because the derivative of the functions is defined at $x=2$ and at $x=-2$.

since the function are continuous, $f_1(2)=f_2(2)$

$-\frac{2^2}{2} +4(2)-2=\frac{(2)^2}{2} +c_1$

$4=2+c$

$\boxed{c_1=2}$

Thus we have

$f_1(x)=\frac{x^2}{2} +2$

Now we can easily evaluate $f(0)$

$f_1(0)=0+2=2\\\\\boxed{f(0)=2}$

Which is choice B.