Math subject GRE test 9768 Q.26

Question

What is the fastest way to solve this question?

I checked and I found that the derivative exists at $x = 1$.

Answer 1

First you can just compute $f$ at $0,1,2$ getting $-2,1,-2$ so clearly A and C are wrong. Because the quadratic term is negative the function will go to $-\infty$ as $x$ gets large in either direction, so it will have some absolute maximum, and D is wrong. You can compute the derivative at $0,2$ and find it is nonzero to rule out $E$. The answer is B. There is no derivative at B because above $1$ is is $-2$ and below $1$ it is $+2$. Since the derivative fails to exist it is a candidate for a local maximum, which it is. It is also the global maximum.

Answer 2

Both components are concave parabolas, with vertices respectively in $x=2$ and $x=0$. So the first component is increasing for $x<1$ and the second component is decreasing for $x \ge 1$. The function is continuous at $x=1$. Therefore (B) is the correct answer.

Answer 3

The functions agree at $1$ so $f$ is continuous. Verify that $f'(x) > 0$ to the left of $1,$ and $f'(x) <0$ to the right of $1.$ It follows that $f$ has an absolute max at $1.$