The absolute maximum value of the function \( f(x)=2x^3-15x^2+36x+7 \) on the interval [1,5] is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 62
On a closed interval, the absolute maximum is the largest of the values at the critical points inside the interval and at the end points.
\( \displaystyle f'(x)=6x^2-30x+36 \)
\( \displaystyle =6(x^2-5x+6) \)
\( \displaystyle =6(x-2)(x-3) \)
The critical points x = 2 and x = 3 lie in [1, 5].
\( \displaystyle f(1)=2-15+36+7=30 \)
\( \displaystyle f(2)=16-60+72+7=35 \)
\( \displaystyle f(3)=54-135+108+7=34 \)
\( \displaystyle f(5)=250-375+180+7=62 \)
The largest value is 62, at the end point x = 5.
Why the other options are incorrect
Option (1) 35 — Incorrect. 35 is the local maximum value, at x = 2; the value at the end point x = 5 is larger.
Option (2) 56 — Incorrect. 56 is not a value of f at any critical point or end point.
Option (4) 30 — Incorrect. 30 is the value at the end point x = 1, which is the absolute minimum here.