Solveeit Logo

Question

Question: The condition that \[f\left( x \right) = a{x^3} + b{x^2} + cx + d\] has no extreme value is A) \[{...

The condition that f(x)=ax3+bx2+cx+df\left( x \right) = a{x^3} + b{x^2} + cx + d has no extreme value is
A) b2>3ac{b^2} > 3ac
B) b2=4ac{b^2} = 4ac
C) b2=3ac{b^2} = 3ac
D) b2<3ac{b^2} < 3ac

Explanation

Solution

Here we are given a function of xx so we will first differentiate it and then put the discriminant of the equation less than zero to get the desired answer.
For a standard quadratic equation ax2+bx+c=0a{x^2} + bx + c = 0
The discriminant is given by:-
D=b24acD = {b^2} - 4ac
If a function has no extrema, it implies that it has no maximum or minimum values.

Complete step by step solution:
We are given that:
f(x)=ax3+bx2+cx+df\left( x \right) = a{x^3} + b{x^2} + cx + d
Differentiating the above function with respect to xx we get:-
ddx[f(x)]=ddx[ax3+bx2+cx+d]\Rightarrow \dfrac{d}{{dx}}\left[ {f\left( x \right)} \right] = \dfrac{d}{{dx}}\left[ {a{x^3} + b{x^2} + cx + d} \right]
ddx[f(x)]=ddx(ax3)+ddx(bx2)+ddx(cx)+ddx(d)\Rightarrow \dfrac{d}{{dx}}\left[ {f\left( x \right)} \right] = \dfrac{d}{{dx}}\left( {a{x^3}} \right) + \dfrac{d}{{dx}}\left( {b{x^2}} \right) + \dfrac{d}{{dx}}\left( {cx} \right) + \dfrac{d}{{dx}}\left( d \right)
ddx[f(x)]=addx(x3)+bddx(x2)+cddx(x)+ddx(d)\Rightarrow \dfrac{d}{{dx}}\left[ {f\left( x \right)} \right] = a\dfrac{d}{{dx}}\left( {{x^3}} \right) + b\dfrac{d}{{dx}}\left( {{x^2}} \right) + c\dfrac{d}{{dx}}\left( x \right) + \dfrac{d}{{dx}}\left( d \right)
Now we know that the derivative of a constant is zero.
Also applying the following formula:
ddx(xn)=nxn1\dfrac{d}{{dx}}\left( {{x^n}} \right) = n{x^{n - 1}}
We get:
ddx[f(x)]=a(3x2)+b(2x)+c(1)+0\dfrac{d}{{dx}}\left[ {f\left( x \right)} \right] = a\left( {3{x^2}} \right) + b\left( {2x} \right) + c\left( 1 \right) + 0
ddx[f(x)]=3ax2+2bx+c\Rightarrow \dfrac{d}{{dx}}\left[ {f\left( x \right)} \right] = 3a{x^2} + 2bx + c
Now, in order to find the extreme values we put
ddx[f(x)]=0\dfrac{d}{{dx}}\left[ {f\left( x \right)} \right] = 0
Therefore we get:-
3ax2+2bx+c=0\Rightarrow 3a{x^2} + 2bx + c = 0
Now, For no extrema, f should have zero turning points, i.e. above quadratic have no real roots.
Therefore, the discriminant of the above equation should be less than zero
Now since for a standard quadratic equation ax2+bx+c=0a{x^2} + bx + c = 0
The discriminant is given by:-
D=b24acD = {b^2} - 4ac
Hence,
Calculating the discriminant for above equation we get:-
D=(2b)24(3a)(c)<0\Rightarrow D = {\left( {2b} \right)^2} - 4\left( {3a} \right)\left( c \right) < 0
4b212ac<0\Rightarrow 4{b^2} - 12ac < 0
On simplification, we get
b23ac<0\Rightarrow {b^2} - 3ac < 0
b2<3ac\Rightarrow {b^2} < 3ac

\therefore Hence, option D is the correct answer.

Note:
In order to find the extema values we need to differentiate the given function and find the values of xx and put that value back in the function and find its value which are the extremum values.
Also, in a quadratic equation if discriminant D is greater than zero (D>0D > 0 ) then its has real and distinct roots.
If discriminant is equal to zero (D=0D = 0 ) then it has equal roots
If discriminant is less than zero (D<0D < 0) then it has imaginary roots.
For a function to have no extrema, the equation of its derivative should have no roots
Therefore the students should put its discriminant to be less than zero in order to get correct answer.