Question

The graph of f ′ (x), the derivative of f(x), is continuous for all x and consists of five line segments as shown below. Given f (5) = 10, find the absolute minimum value of f (x) over the interval [0, 5].
A.3
B.0
C.7
D.17

Answer 1

According to the graph, we have for $0$

$f'(x)=\begin{cases}x&\text{for }0$

$f'$ is positive for all $0$, which means $f$ is strictly increasing on the interval [0, 5], and so its minimum value on this interval occurs at $x=0$.

We can integrate $f'$ along each subinterval above to find $f$:

$f(x)=\begin{cases}\dfrac{x^2}2+C_1&\text{for }0\le x\le 2\\\\2x+C_2&\text{for }2\le x\le 4\\\\10x-x^2+C_3&\text{for }4\le x\le5\end{cases}$

$f'$ is continuous, which means $f$ is also continuous. This means

$\displaystyle\lim_{x\to2}\left(\dfrac{x^2}2+C_1\right)=\lim_{x\to2}(2x+C_2)$

$\implies2+C_1=4+C_2\implies C_1=2+C_2$

and

$\displaystyle\lim_{x\to4}(2x+C_2)=\lim_{x\to4}(10x-x^2+C_3)$

$\implies8+C_2=24+C_3\implies C_2=16+C_3$

Given that $f(5)=10$, we have

$10(5)-5^2+C_3=10\implies C_3=-15\implies C_2=1\implies C_1=3$

so that over $0\le x\le 2$ we have $f(x)=\dfrac{x^2}2+3$. We know the minimum value occurs at $x=0$, so it is 3 and the answer is A.