Question

The image of the interval [-1, 3] under the mapping $f:R\to R$ given by $f\left( x \right)=4{{x}^{3}}-12x$ is
\[\begin{align}
  & A.\left[ 8,72 \right] \\
 & B.\left[ 0,72 \right] \\
 & C.\left[ 0,8 \right] \\
 & D.\left[ -8,8 \right] \\
\end{align}\]

Answer 1

Hint: To solve this question, we will first of all compute f'(x) and put \[\text{f}'\left( \text{x} \right)=0\]. After doing so, we will try to obtain all values of x possible when f'(x) becomes equal to 0. Finally, we will take the value of f(x) for all x values and compute which one is greatest or least to compute the interest.

Complete step by step answer:
Least or largest values of a function g(x), $g\left( x \right):X\to Y$ can be obtained by computing g'(x) and putting \[\text{g}'\left( \text{x} \right)=0\] to get values of x and then finally compute g at that obtained value of x to get highest and lowest values. So, we will use this technique for f(x).
We have $f\left( x \right)=4{{x}^{3}}-12x$ and interval is given as [-1, 3], when x = -1.
\[\begin{align}
  & f\left( x \right)=4{{x}^{3}}-12x \\
 & f\left( -1 \right)=4{{\left( -1 \right)}^{3}}-12\left( -1 \right) \\
 & \Rightarrow 4\left( -1 \right)+12 \\
 & \Rightarrow -4+12 \\
 & \Rightarrow 8 \\
\end{align}\]
So, \[f\left( -1 \right)=8\text{ }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. (i)}\]
When x = 3 then f(x) is
\[\begin{align}
  & f\left( x \right)=4{{x}^{3}}-12x \\
 & f\left( 3 \right)=4{{\left( 3 \right)}^{3}}-12\left( 3 \right) \\
 & \Rightarrow 4\times 27-36 \\
 & \Rightarrow 108-36 \\
 & \Rightarrow 72 \\
\end{align}\]
So, \[f\left( 3 \right)=72\text{ }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. (ii)}\]
Now, finally we will compute f'(x).
Differentiating f(x) with respect to x, we get:
\[f'\left( x \right)=\dfrac{d}{dx}\left( 4{{x}^{3}} \right)-\dfrac{d}{dx}\left( 12x \right)\]
$\dfrac{d}{dx}\left( {{x}^{n}} \right)=n{{x}^{n-1}}$ using this we have:
\[\begin{align}
  & f'\left( x \right)=4\times 3{{x}^{2}}-12 \\
 & f'\left( x \right)=12{{x}^{2}}-12 \\
\end{align}\]
Putting \[\text{f}'\left( \text{x} \right)=0\] we get:
\[\begin{align}
  & f'\left( x \right)=0 \\
 & 12{{x}^{2}}-12=0 \\
 & 12{{x}^{2}}=12 \\
\end{align}\]
Cancelling 12 from both sides, we get:
\[\begin{align}
  & {{x}^{2}}=1 \\
 & x=+1,-1 \\
\end{align}\]
So, the possible values of x are +1 and -1.
$f\left( x \right)=f\left( -1 \right)=8$ already computed and $f\left( x \right)=f\left( 1 \right)=4\times 1-12\times 1=4-12=-8\text{ }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. (iii)}$
So from equation (i), (ii) and (iii) we get that max value of f(x)=72 at x=3 and minimum value of f(x)=-8 at x=1.
The image interval is given by [-8, 72] so option C is correct.

Note:
The key point to note here in this question is that the maximum value of f(x) as 72 is obtained at x=3 and not on $x\ne 1$ So while calculating least or greatest value apart from all x obtained by putting f'(x)=0 also check for all x which comes in given interval of the question.