Question

The exhaustive set of values of \[m\] for which \[f(x) = \dfrac{{{x^3}}}{3} + (m - 1){x^2} + (m + 5)x + 7\] is increasing function \[\forall x > 0\] is \[[K,\infty )\] then the value of \[K\] is

Answer 1

Hint: The derivative of a function may be used to determine whether the function is increasing or decreasing on any intervals in its domain. If \[f'(x) > 0\] at each point in an interval I, then the function is said to be increasing on I. \[f'(x) < 0\] at each point in an interval I, then the function is said to be decreasing on I.

Complete step-by-step answer:
If \[f'(x) > 0\] then \[f\] is increasing on the interval, and if \[f'(x) < 0\] then \[f\] is decreasing on the interval.
The steps involved in the process of finding the intervals of increasing and decreasing function, are as follows:
Firstly, differentiate the given function with respect to the constant variable.
Then solve \[f'(x) = 0\] .
After solving the equation of the first derivative and finding the points of discontinuity we get the open intervals with the value of \[x\] , through which the sign of the intervals can be taken into consideration.
If the sign of the interval in their first derivative form gives more than \[0\] then the function is said to be increasing in nature, while if the sign of the intervals in their first derivative form gives less than \[0\] then the function is said to be decreasing in nature.
Finally, we get increasing as well as decreasing intervals of the function.
We are given the function \[f(x) = \dfrac{{{x^3}}}{3} + (m - 1){x^2} + (m + 5)x + 7\]
On differentiating the function with respect to \[x\] we get ,
 \[f'(x) = {x^2} + 2(m - 1)x + (m + 5)\]
Given \[f(x)\] is increasing \[\forall x > 0\]
 \[m - 1 > 0\] and \[m + 5 > 0\]
Therefore \[m > 1\]
Hence we get \[K = 1\] .
So, the correct answer is “K = 1”.

Note: In determining intervals where a function is increasing or decreasing, you first find domain values where all critical points will occur; then, test all intervals in the domain of the function of these values to determine if the derivative is positive or negative.