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

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

Given, f(x) = x^3 – 12x^2 + 45x + 8. What is the minimum value of f(x)?
 
(a) -1
 
(b) 0
 
(c) 1
 
(d) Value does not exist
 
I have been asked this question during an online exam.
 
Enquiry 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

Right answer is (c) 1
 
The 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’(5 – h) = 3*(5 – h – 3)(5 – h – 5) = -3h(2 – h) = negative.
 
f’(5 + h) = 3*(5 + h – 3)(5 + h – 5) = 3h(2 + h) = positive.
 
Clearly, f’(x) changes sign from negative on the left to positive on the right of the point x = 5.
 
So, f(x) has minimum at 5.
 
Putting, x = 5 in (1)
 
Thus, its maximum value is,
 
f(3) = 5^3 – 12*5^2 + 45*5 + 8 = 58.