Question

If $f(x) = 3{x^3} - 9{x^2} - 27x + 15$, then the maximum value of \[f\left( x \right)\] is.
A. -66
B. 30
C. -30
D. 66

Answer 1

Hint:
To find the maximum value of $f(x)$ we need to first find the points of maxima and minima of the curve given to us, this is done by differentiating $f(x)$ by x and putting $f'(x)$ equals to zero. The solutions to the formed equation are the points of maxima and minima. Now we can either put all the points of extremities one by one and find the maximum value or we can do a double derivative and find the points of maxima, if the value is smaller than 0, then only it is a point of maxima else it would be minima. After that, we will put the maxima values in $f(x)$ to get the maximum value of $f(x)$ .

Complete step by step solution:
We are given
 $f(x) = 3{x^3} - 9{x^2} - 27x + 15$ … (1)
Now, to find points of maxima and minima, we have to differentiate $f(x)$ with respect to x and then equate the differentiated equation to 0 to find the points of maxima and minima.
To differentiate (1) we need to know the following differentiation formulas:
 $\dfrac{{d{x^n}}}{{dx}} = n \times {x^{n - 1}}$ … (2)
 $\dfrac{\text{ Constant}}{{dx}} = 0$ … (3)
Hence, using (2) and (3) to differentiate (1) with respect to x, we get
 $ \Rightarrow f'(x) = 9{x^2} - 18x - 27$ … (4)
Now, put $f'(x) = 0$ , we get
 $ \Rightarrow 9{x^2} - 18x - 27 = 0$
Dividing the whole equation by 9, we get
 $ \Rightarrow {x^2} - 2x - 3 = 0$
Now to solve the above quadratic equation we should know Middle term splitting, in middle term splitting if the quadratic equation is given by $a{x^2} + bx + c = 0$ , then we have to split the middle term ‘b’ such that the two split terms are equal to the product of ‘a’ and ‘c’. After that, we take the common terms to further factorize it and reach the answer.
Solving equation by middle term split method, we get
We can split the term -2 in -3 and +1, after splitting, we get
 $ \Rightarrow {x^2} + x - 3x - 3 = 0$
Taking relevant common terms, we get
 $ \Rightarrow x(x + 1) - 3(x + 1) = 0$
Now taking $(x + 1)$ common, we get
 $ \Rightarrow (x + 1)(x - 3) = 0$
Hence the solutions to the above equation can be 3, -1
These are the points of maxima and minima.
Now to find which are maxima and minima we need to find the double derivative of \[f\left( x \right)\] and put the values of points of maxima and minima we got (i.e. 3 and -1). If , it is a point of minima else it is a point of maxima.
Hence, differentiating (4), we get
 $ \Rightarrow f''(x) = 18x - 18$
Taking 18 common, we get
 $ \Rightarrow f''(x) = 18(x - 1)$ … (5)
Put 3 in (5), we get
 $ \Rightarrow f''(3) = 18(3 - 1) = 36$
Put -1 in (5), we get
 $ \Rightarrow f''( - 1) = 18( - 1 - 1) = - 36$
Clearly,
 $ \Rightarrow f''(3) > 0$
Hence it is a point of minima
And,
 $ \Rightarrow f''( - 1) < 0$
Hence it is a point of maxima
Now, to find the maximum value of $f(x)$ , we have to put point(s) of maxima in $f(x)$ ,
Since we have one point of maxima (-1), we have to find $f( - 1)$ ,
 $ \Rightarrow f( - 1) = 3 \times {( - 1)^3} - 9 \times {( - 1)^2} - 27 \times ( - 1) + 15$
After calculating RHS, we get
 $ \Rightarrow f( - 1) = 30$

Hence the correct option is B.

Note:
Making wise decisions to save time is the key to winning the competition, for instance in the above example, we can simply avoid the double differentiation because we have only two points of maxima and minima. It would be easier for us to find $f(x)$ and $f'(x)$ rather than doing the double differentiation. But, the double differentiation comes into play when there are many points of maxima and minima, so it is recommended to use double differentiation only when needed.