Question

Read the following passage and answer the questions given below.

The temperature of a person during an intestinal illness is given by f(x) = 0.1x2 + mx + 98.6, 0 ≤ x ≤ 12, m being a constant, where f(x) is the temperature in °F at x days.

1. Is the function differentiable in the interval (0, 12)? Justify your answer.
2. If 6 is the critical point of the function, then find the value of the constant m.
3. Find the intervals in which the function is strictly increasing/strictly decreasing.
   OR
   Find the points of local maximum/local minimum, if any, in the interval (0, 12) as well as the points of absolute maximum/absolute minimum in the interval [0, 12]. Also, find the corresponding local maximum/local minimum and the absolute ‘maximum/absolute minimum values of the function.

Answer 1

i. f(x) = –0.1x² + mx + 98.6, being a polynomial function, is differentiable everywhere, hence, differentiable in (0, 12)

ii. f'(x) = –0.2x + m

Since, 6 is the critical point,

f'(6) = 0

⇒ m = 1.2

iii. f(x) = –0.1x² + 1.2x + 98.6

f'(x) = –0.2x + 1.2 = –0.2(x – 6)

In the Interval	f'(x)	Conclusion
(0, 6)	+ve	f is strictly increasing in [0, 6]
(6, 12)	-ve	f is strictly decreasing in [6, 12]

OR
iii. f(x) = –0.1x² + mx + 98.6,

f'(x) = –0.2x + 1.2, f'(6) = 0,

f"(x) = –0.2

f"(6) = –0.2 < 0

Hence, by second derivative test 6 is a point of local maximum. The local maximum value = f(6) = −0.1 × 6² + 1.2 × 6 + 98.6 = 102.2

We have f(0) = 98.6, f(6) = 102.2, f(12) = 98.6

6 is the point of absolute maximum and the absolute maximum value of the function = 102.2.

0 and 12 both are the points of absolute minimum and the absolute minimum value of the function = 98.6.