Question

The function $f$ defined by $f\left( x \right) = 4{x^4} - 2x + 1$ in increasing for
A.$x < 1$
B.$x > 0$
C.$x < \dfrac{1}{2}$
D.$x > \dfrac{1}{2}$

Answer 1

Hint : For the function to be increasing we take the first derivative test of the given function and then put it greater than zero , and then find the value of $x$ for the function to be increasing . To check the function to be increasing, put the different values of $x$ in the given function and draw the graph .

Complete step-by-step answer :
Given : $f\left( x \right) = 4{x^4} - 2x + 1$
Now differentiating the function for the first derivative test we have
${f^1}\left( x \right) = 16{x^3} - 2$ , on simplifying we get ,
${f^1}\left( x \right) = 2\left( {8{x^3} - 1} \right)$ .
Now putting the first derivative greater than zero , we have
${f^1}\left( x \right) > 0$
On putting the values we get
$= 2\left( {8{x^3} - 1} \right) > 0$ , on solving we get ,
$= \left( {8{x^3} - 1} \right) > 0$
On further solving we get
$= 8{x^3} > 1$ , on simplifying we get ,
$= {x^3} > \dfrac{1}{8}$
Now taking the cube root on both sides we get ,
$= x > \dfrac{1}{2}$ .
Therefore , at $x > \dfrac{1}{2}$ the function will be increasing .
Therefore , option ( D ) is the correct answer for the given question .
So, the correct answer is “Option D”.

Note : The derivative of a function is used to determine whether the function is increasing or decreasing on any intervals in its domain . If ${f^1}\left( x \right) > 0$ at each point in an interval $I$, then the function is said to be increasing on I. ${f^1}\left( x \right) > 0$ at each point in an interval I, then the function is said to be decreasing on $I$ , Because the derivative is zero or does not exist only at critical points of the function, it must be positive or negative at all other points where the function exists . In determining intervals where a function is increasing or decreasing, you first find domain values where all critical points will occur .