Question

The image of the interval \[\left[ { - 1,{\rm{ }}3} \right]\;\] under the mapping \[f\left( x \right) = 4{x^2} - 12x\;\] is
A) \[\left[ { - 9,{\rm{ 0}}} \right]\;\]
B) \[\left[ { - 9,{\rm{ 16}}} \right]\;\]
C) \[\left[ { - 8,{\rm{ 0}}} \right]\;\]
D) None of these

Answer 1

Hint:
To solve this problem, we will first differentiate the given function with respect to the given variable and then we will equate this with zero. From there, we will find all possible values of the variable. Then we will check the maximum and minimum values by putting them in the given function. From there, we will get the required image of the given interval.

Complete step by step solution:
We know that the least and greatest values of a function say \[g\left( x \right)\] can be obtained by computing the \[g'\left( x \right)\] and putting \[g'\left( x \right) = 0\] to get all possible values of \[x\] . So we will use this technique for \[f\left( x \right)\].
We have \[f\left( x \right) = 4{x^2} - 12x\;\]and the interval is given as \[\left[ { - 1,{\rm{ }}3} \right]\;\]
We will compute the value of \[f\left( x \right)\] when \[x = - 1\].
\[ \Rightarrow f\left( { - 1} \right) = 4{\left( { - 1} \right)^2} - 12\left( { - 1} \right)\;\]
On multiplying the terms, we get
\[ \Rightarrow f\left( { - 1} \right) = 4 + 12\]
On adding these numbers, we get
\[ \Rightarrow f\left( { - 1} \right) = 16\] …………. \[\left( 1 \right)\]
So, we have got the value \[f\left( { - 1} \right) = 16\]
Similarly, We will compute the value of \[f\left( x \right)\] when \[x = 3\].
\[ \Rightarrow f\left( 3 \right) = 4{\left( 3 \right)^2} - 12\left( 3 \right)\;\]
On multiplying the terms, we get
\[ \Rightarrow f\left( 3 \right) = 36 - 36\]
On subtracting these numbers, we get
\[ \Rightarrow f\left( 3 \right) = 0\] ……… \[\left( 2 \right)\]
So, we have got the value \[f\left( 3 \right) = 0\]
Now, we will compute \[f'\left( x \right)\].
\[ \Rightarrow f'\left( x \right) = \dfrac{d}{{dx}}\left( {4{x^2} - 12x} \right)\;\]
On differentiating each term, we get
\[ \Rightarrow f'\left( x \right) = 8x - 12\;\]
Now, we will put \[f'\left( x \right) = 0\]
\[ \Rightarrow f'\left( x \right) = 8x - 12\; = 0\]
On further simplification, we get
\[ \Rightarrow 8x\; = 12\]
Dividing both sides by 8, we get
\[ \Rightarrow x\; = \dfrac{{12}}{8}\]
On further simplifying the fraction, we get
\[ \Rightarrow x\; = \dfrac{3}{2}\]
Now, we will compute the value of \[f\left( x \right)\] when \[x = \dfrac{3}{2}\].
\[ \Rightarrow f\left( {\dfrac{3}{2}} \right) = 4{\left( {\dfrac{3}{2}} \right)^2} - 12\left( {\dfrac{3}{2}} \right)\;\]
On multiplying the terms, we get
\[ \Rightarrow f\left( {\dfrac{3}{2}} \right) = 9 - 18\]
On subtracting these numbers, we get
\[ \Rightarrow f\left( {\dfrac{3}{2}} \right) = - 9\] …………. \[\left( 3 \right)\]
From equation 1, equation 2 and equation 3, we have got the maximum value of \[f\left( x \right)\] as 16 at \[x = - 1\] and minimum value of \[f\left( x \right)\] as -9 at \[x = \dfrac{3}{2}\].
Thus, the image interval is given by \[\left[ { - 9,{\rm{ 16}}} \right]\;\].

Hence, the correct option is option B.

Note:
Since we have obtained the maximum and minimum value of a function. Here, the maximum or minimum of the function are known as "Global" or "Absolute" maximum or minimum. There can be only one global maximum (and one global minimum) but there can be more than one local minimum or maximum.