FORM GR1268 Question 7

Question:

The figure above shows the graph of the derivative \(f'\) of a function \(f\), where \(f\) is continuous on the interval \([0,4]\) and differentiable on the interval \((0,4)\). Which of the following gives the correct ordering of the values \(f(0), f(2),\) and \(f(4)\)?

(A) \(f(0)\lt f(2)\lt f(4)\)

(B) \(f(0)\lt f(4)=f(2)\)

(C) \(f(0)\lt f(4)\lt f(2)\)

(D) \(f(4)=f(2)\lt f(0)\)

(E) \(f(4)\lt f(0)\lt f(2)\)

Answer:

(C)

Answer Key:

Since \(f'(x)\gt0\) for \(0\lt x\lt 2\), \(f\) is increasing on the interval \((0,2)\).

Since \(f'(x)\lt0\) for \(2\lt x\lt 4\), \(f\) is decreasing on the interval \((2,4)\).

So, \(f(2)\) is the largest.

Now, the question is, which is larger, \(f(0)\) or \(f(4)\)?

Intuitively, since the rate of increase on the interval \((0,2)\) is larger than the rate of decrease on the interval \((2,4)\), \(f(0)\) is "further away" from \(f(2)\).

Hence, \(f(0)\lt f(4)\).

Formally, since \(\displaystyle\int_0^2 f'(x)dx=f(2)-f(0)\), we have \(\displaystyle f(0)=f(2)-\int_0^2f'(x)dx\).

Similarly, since \(\displaystyle\int_2^4f'(x)dx=f(4)-f(2)\), we have \(\displaystyle f(4)=f(2)+\int_2^4f'(x)dx\).

Because \(\displaystyle\int_0^2f'(x)dx\gt -\int_2^4f'(x)dx\), we have \(f(0)\lt f(4)\).