Question

The least value of the function f(x) =\[{x^3} - 18{x^2} + 96x\] in the interval [0, 9] is
A. 126
B. 135
C. 160
D. 0

Answer 1

Hint: In this question we will use the concept of maxima and minima. First of all, we will differentiate the given function and equate it to zero to find the critical points and finally calculate the value of the function at the critical points and endpoints to get the minimum value of the function.

Complete step-by-step solution:
It is given, f(x) = \[{x^3} - 18{x^2} + 96x\] and we have to find the least value.
We know that the least or maximum value of a function occurs either at the critical points or at the endpoints. so, we will first calculate the critical points.
To find the critical points, we will differentiate the given function and then equate it to zero.
f(x) =\[{x^3} - 18{x^2} + 96x\].
$\therefore f’(x)$ = \[3{x^2} - 36x + 96\]
Equating $ f’(x)$ to zero, we have:
 \[3{x^2} - 36x + 96\]=0
$ \Rightarrow 3(x – 8)(x – 4)=0$
$ \Rightarrow x = 4,8 $.
Now check the value of f(x) by putting value ‘x’ one by one.
Where $x = 0,4,8,9.$
When $x=0:$
f(x) = \[{0^3} - 18 \times {0^2} + 96 \times 0 = 0\]
When $x = 4:$
f(x) = \[{4^3} - 18 \times {4^2} + 96 \times 4 =260\]
When $x = 8:$
f(x) = \[{8^3} - 18 \times {8^2} + 96 \times 8 = 120\]
When $x = 9:$
f(x) = \[{9^3} - 18 \times {9^2} + 96 \times 9 = 135\]
Therefore, the least value is $0$
Hence, the option D is correct.

Note: In this type of question, you should have a clear idea of critical points. The critical points are the value of ‘x’ where the function is not differentiable or the derivative of the function is zero. The same process that we have used in this question can be used to calculate the maximum value of the function.