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Question: The image of the interval [-1, 3] under the mapping \[f\left( x \right) = 4x{^3} - 12x\] is A) [-...

The image of the interval [-1, 3] under the mapping f(x)=4x3−12xf\left( x \right) = 4x{^3} - 12x is
A) [-2, 0]
B) [-8, 72]
C) [-8, 0]
D) [-8, -2]

Explanation

Solution

Hint : The equation of the curve and the interval of these curves is also given. We can substitute the values lying between the given intervals in the function and see the minimum and maximum values obtained. These value intervals will provide us the required image interval.

Complete step by step solution:
The given equation of the curve is f(x)=4x3−12xf\left( x \right) = 4x{^3} - 12x and its interval is [-1, 3].
The image of this interval under the mapping of the given curve will be given by the minimum and maximum value of this function when x belongs to this interval.
We have 5 values of x lying between the intervals that are: -1, 0, 1, 2, 3. Substituting these values of variable in the given function, we get:
When x = -1:

f(−1)=4(−1)3−12(−1) ⇒f(−1)=−4+12 ⇒f(−1)=8   f\left( { - 1} \right) = 4\left( { - 1} \right){^3} - 12\left( { - 1} \right) \\\ \Rightarrow f\left( { - 1} \right) = - 4 + 12 \\\ \Rightarrow f\left( { - 1} \right) = 8 \;

When x = 0:

f(0)=4(0)3−12(0) ⇒f(0)=0−0 ⇒f(0)=0   f\left( 0 \right) = 4\left( 0 \right){^3} - 12\left( 0 \right) \\\ \Rightarrow f\left( 0 \right) = 0 - 0 \\\ \Rightarrow f\left( 0 \right) = 0 \;

When x = 1:

f(1)=4(1)3−12(1) ⇒f(1)=4−12 ⇒f(1)=−8   f\left( 1 \right) = 4\left( 1 \right){^3} - 12\left( 1 \right) \\\ \Rightarrow f\left( 1 \right) = 4 - 12 \\\ \Rightarrow f\left( 1 \right) = - 8 \;

When x = 2:

f(2)=4(2)3−12(2) ⇒f(2)=32−24 ⇒f(2)=8   f\left( 2 \right) = 4\left( 2 \right){^3} - 12\left( 2 \right) \\\ \Rightarrow f\left( 2 \right) = 32 - 24 \\\ \Rightarrow f\left( 2 \right) = 8 \;

When x = 3:

f(3)=4(3)3−12(3) ⇒f(3)=108−36 ⇒f(3)=72   f\left( 3 \right) = 4\left( 3 \right){^3} - 12\left( 3 \right) \\\ \Rightarrow f\left( 3 \right) = 108 - 36 \\\ \Rightarrow f\left( 3 \right) = 72 \;

We can see that the minimum and maximum values of the function in this interval is -8 and 72 respectively.
So, the image of the interval for the given curve is [-8, 72]
So, the correct answer is “Option B”.

Note : We can also find the required image by the derivative method.
The given curve is f(x)=4x3−12xf\left( x \right) = 4x{^3} - 12x
The derivative of this function is given as:

f′(x)=12x2−12x (∵f′(x)n)=nxn−1 ⇒f′(x)=12(x2−1)  f'\left( x \right) = 12x{^2} - 12x \\\ \left( {\because f'{{\left( x \right)}^n}} \right) = n{x^{n - 1}} \\\ \Rightarrow f'\left( x \right) = 12\left( {x{^2} - 1} \right) \\\

When we equate this equal to 0, we get the value as:
f′(x)=0 ⇒12(x2−1)=0 ⇒x2−1=0 ⇒x2=1 ⇒x=±1   f'(x) = 0 \\\ \Rightarrow 12\left( {{x^2} - 1} \right) = 0 \\\ \Rightarrow {x^2} - 1 = 0 \\\ \Rightarrow {x^2} = 1 \\\ \Rightarrow x = \pm 1 \;
The minimum value will be given by x>0x > 0 i.e. when x = 1.
⇒f(1)=4(1)3−12(1) ⇒f(1)=−8   \Rightarrow f\left( 1 \right) = 4{\left( 1 \right)^3} - 12\left( 1 \right) \\\ \Rightarrow f\left( 1 \right) = - 8 \;
Now, the interval extremes are -1 and 3 and either of them can give the maximum value, so we can try for both.

f(−1)=4(−1)3−12(−1) ⇒f(−1)=8   f\left( { - 1} \right) = 4\left( { - 1} \right){^3} - 12\left( { - 1} \right) \\\ \Rightarrow f\left( { - 1} \right) = 8 \; f(3)=4(3)3−12(3) ⇒f(3)=72   f\left( 3 \right) = 4\left( 3 \right){^3} - 12\left( 3 \right) \\\ \Rightarrow f\left( 3 \right) = 72 \;

So, the maximum value is 8.
Therefore, the image will be at [-8, 72]