Question
Question: The function \[f\left( x \right) = {a^2}\dfrac{{{x^3}}}{3} + a{x^2} + x\] is strictly increasing for...
The function f(x)=a23x3+ax2+x is strictly increasing for all real x, if
A) a=0
B) a=x1
C) a=−x1
D) None of these
Solution
To solve the question, we will differentiate the function to find whether the function is increasing or decreasing in a given domain. Once we obtain the trend of the function we can use the given condition to find which one is the suitable option for the question.
Complete step by step solution:
We have, f(x)=a23x3+ax2+x
Differentiating the function with respect to x, we have,
⇒f′(x)=dxd(a23x3+ax2+x)
⇒f′(x)=dxd(a23x3)+dxd(ax2)+dxd(x)
Taking constants outside the differentials, we get,
⇒f′(x)=3a2dxd(x3)+adxd(x2)+dxd(x)
Now, we know the power rule of differentiation as dxd(xn)=nxn−1. So, we get,
⇒f′(x)=3a2×(3x2)+a×(2x)+(1)
Opening the brackets, we get,
⇒f′(x)=a2x2+2ax+1
Now, condensing the terms into a whole square using the algebraic identity (a+b)2=a2+2ab+b2. So, we get,
⇒f′(x)=(ax+1)2
Now, we know that the square of any entity is always non negative. So, we have, f′(x)⩾0.
For the function to be strictly increasing, f′(x)>0.
So, the derivative of the function, f′(x) should not be equal to zero.
Thus, f′(x)=(ax+1)2=0
⇒ax+1=0
⇒a=−x1
Therefore, the correct option is (C).
Note:
The first derivative test of a function tells us about the increasing and decreasing nature of the function. If the first derivative of a function is positive, then the function is strictly increasing. We must read the question carefully and pay attention to words like ‘strictly increasing’ as we don’t include the case of derivative being equal to zero.