Given, f(x) = x^3 – 12x^2 + 45x + 8. At which point does f(x) has its maximum?

Category: QuestionsGiven, f(x) = x^3 – 12x^2 + 45x + 8. At which point does f(x) has its maximum?
Editor">Editor Staff asked 11 months ago

Given, f(x) = x^3 – 12x^2 + 45x + 8. At which point does f(x) has its maximum?
 
(a) 1
 
(b) 2
 
(c) 3
 
(d) 4
 
I had been asked this question in exam.
 
This question is from Calculus Application in section Application of Calculus of Mathematics – Class 12
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1 Answers
Editor">Editor Staff answered 11 months ago

Correct answer is (c) 3
 
Best explanation: We have, f(x) = x^3 – 12x^2 + 45x + 8  ……….(1)
 
Differentiating both sides of (1) with respect to x we
 
f’(x) = 3x^2 – 24x + 45
 
3x^2 – 24x + 45 = 0
 
Or x^2 – 8x + 15 = 0
 
Or (x – 3)(x – 5) = 0
 
Thus, either x – 3 = 0 i.e., x = 3 or x – 5 = 0 i.e., x = 5
 
Therefore, f’(x) = 0 for x = 3 and x = 5.
 
If h be a positive quantity, however small, then,
 
f’(3 – h) = 3*(3 – h – 3)(3 – h – 5) = 3h(h + 2) = positive.
 
f’(3 + h) = 3*(3 + h – 3)(3 + h – 5) = 3h(h – 2) = negative.
 
Clearly, f’(x) changes sign from positive on the left to negative on the right of the point x = 3.
 
So, f(x) has maximum at 3.