Given, f(x) = x^3 – 12x^2 + 45x + 8. What is the maximum value of f(x)?

Category: QuestionsGiven, f(x) = x^3 – 12x^2 + 45x + 8. What is the maximum value of f(x)?
Editor">Editor Staff asked 11 months ago

Given, f(x) = x^3 – 12x^2 + 45x + 8. What is the maximum value of f(x)?
 
(a) 61
 
(b) 62
 
(c) 63
 
(d) 54
 
The question was asked during an interview.
 
Origin of the question is Calculus Application topic 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 (b) 62
 
For explanation I would say: 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.
 
Putting, x = 3 in (1)
 
Thus, its maximum value is,
 
f(3) = 3^3 – 12*3^2 + 45*3 + 8 = 62.