Question

The set of values of x for which $f\left( x \right) = {x^3} - 6{x^2} + 27x + 10$ is increasing in:
(A) $\left( {1,2} \right)$
(B) $\left( { - \inf ,1} \right) \cup \left( {2,\inf } \right)$
(C) $\left( { - \inf ,\inf } \right)$
(D) $\left( { - \inf ,1} \right)$

Answer 1

Hint : The given problem revolves around the graph of curve, its slope and nature. It involves the concepts of applications of derivatives. The increasing and decreasing nature of the curve is judged by its first derivative test as it gives an idea about the slope and the nature of the curve. The nature of the function is increasing in the intervals of values of x where the derivative of the function is positive and the function is decreasing in intervals of values of x where the derivative of the function is negative.

Complete step by step solution:
In the problem, we are given the function $f\left( x \right) = {x^3} - 6{x^2} + 27x + 10$ .
So, to find the increasing or decreasing nature of a curve, we have to find the variation of slope of the curve. To get an idea of the slope of the curve, we need to find the first order derivative of f(x).
So, $f\left( x \right) = {x^3} - 6{x^2} + 27x + 10$
Differentiating both sides with respect to x, we get,
$f'(x) = 3{x^2} - 12x + 27$
Now, we factorise and simplify the first derivative of the function. So, we get,
$\Rightarrow f'(x) = 3\left( {{x^2} - 4x + 9} \right)$
Now, we have to find the values of x for which the given function $f\left( x \right) = {x^3} - 6{x^2} + 27x + 10$ is increasing. So, we find the intervals of the variable in which the first order derivative $f'(x) = 3\left( {{x^2} - 4x + 9} \right)$ is positive.
For the first order derivative to be positive, we get,
$\Rightarrow f'(x) = 3\left( {{x^2} - 4x + 9} \right) > 0$
Dividing both sides of the inequality by $3$ , we get,
$\Rightarrow \left( {{x^2} - 4x + 9} \right) > 0$
Now, we complete the square on the left side of the inequality.
$\Rightarrow {x^2} - 2\left( 2 \right)x + {2^2} + 5 > 0$
Using the algebraic identity ${a^2} - 2ab + {b^2} = {\left( {a - b} \right)^2}$ , we get,
$\Rightarrow {\left( {x - 2} \right)^2} + 5 > 0$
Now, we know that the square of any real number is always positive. Also, the sum of two positive numbers is always positive. Hence, the above inequality holds true for every value of x. Therefore, the first order derivative $f'(x) = 3\left( {{x^2} - 4x + 9} \right)$ is always positive.
Hence, the set of values of x for which the function $f\left( x \right) = {x^3} - 6{x^2} + 27x + 10$ is increasing is $\left( { - \inf ,\inf } \right)$ . Thus, option (C) is correct.
So, the correct answer is “Option C”.

Note : We must know the relation of the increasing or decreasing nature of the curve to the applications of derivatives. The first order derivative of a function tells us about the slope of the curve and the increasing or decreasing nature of a function. Second order derivative tells us about the critical points at which the function attains its maximum and minimum value. We must know simplification and transposition rules in order to solve the inequality formed while solving the problem.