Question

The condition for \[y = a{x^4} + b{x^3} + c{x^2} + dx + e\] to have points of inflection is
(a) \[{b^2} - 4ac > 0\]
(b) \[3{b^2} - 8ac = 0\]
(c) \[3{b^2} - 8ac > 0\]
(d) \[3{b^2} - 8ac < 0\]

Answer 1

Hint:
Here, we need to find the condition for which the given equation has points of inflection. A function \[f\left( x \right)\] has points of inflection if its second derivative is equal to zero, that is \[f''\left( x \right) = 0\]. First, we will find the second derivative of the given function. Then, we will solve the resulting quadratic function using the quadratic formula and use the expression with the determinant to find the required condition.
Formula Used:
1) The derivative of a function of the form \[af\left( x \right) + b\] can be written as \[\dfrac{{d\left( {af\left( x \right) + b} \right)}}{{dx}} = a\dfrac{{d\left( {f\left( x \right)} \right)}}{x} + b\].
2) The derivative of the function of the form \[f\left( x \right) = {x^n}\] is written as \[f'\left( x \right) = n{x^{n - 1}}\].
3)The derivative of a constant is always 0.
4) The quadratic formula states that the roots of a quadratic equation \[A{x^2} + Bx + C = 0\] are given by \[x = \dfrac{{ - B \pm \sqrt D }}{{2A}}\], where \[D\] is the discriminant given by the formula \[D = {B^2} - 4AC\].

Complete step by step solution:
A function \[f\left( x \right)\] has points of inflection if its second derivative is equal to zero, that is \[f''\left( x \right) = 0\].
First, we will find the first derivative of the given function.
Differentiating both sides of the equation with respect to \[x\], we get
\[\dfrac{{dy}}{{dx}} = \dfrac{{d\left( {a{x^4} + b{x^3} + c{x^2} + dx + e} \right)}}{{dx}}\]
The right hand side is the derivative of a sum of functions.
Therefore, we can simplify the equation to get
\[\dfrac{{dy}}{{dx}} = \dfrac{{d\left( {a{x^4}} \right)}}{{dx}} + \dfrac{{d\left( {b{x^3}} \right)}}{{dx}} + \dfrac{{d\left( {c{x^2}} \right)}}{{dx}} + \dfrac{{d\left( {dx} \right)}}{{dx}} + \dfrac{{d\left( e \right)}}{{dx}}\]
Therefore, the equation becomes
We know that the derivative of a function of the form \[af\left( x \right) + b\] can be written as \[\dfrac{{d\left( {af\left( x \right) + b} \right)}}{{dx}} = a\dfrac{{d\left( {f\left( x \right)} \right)}}{x} + b\].
Therefore, we get
\[ \Rightarrow \dfrac{{d\left( {a{x^4}} \right)}}{{dx}} = a\dfrac{{d\left( {{x^4}} \right)}}{{dx}}\]
\[ \Rightarrow \dfrac{{d\left( {b{x^3}} \right)}}{{dx}} = b\dfrac{{d\left( {{x^3}} \right)}}{{dx}}\]
\[ \Rightarrow \dfrac{{d\left( {c{x^2}} \right)}}{{dx}} = c\dfrac{{d\left( {{x^2}} \right)}}{{dx}}\]
\[ \Rightarrow \dfrac{{d\left( {dx} \right)}}{{dx}} = d\dfrac{{d\left( x \right)}}{{dx}}\]
\[ \Rightarrow \dfrac{{d\left( e \right)}}{{dx}} = e\dfrac{{d\left( 1 \right)}}{{dx}}\]
Substituting \[\dfrac{{d\left( {a{x^4}} \right)}}{{dx}} = a\dfrac{{d\left( {{x^4}} \right)}}{{dx}}\], \[\dfrac{{d\left( {b{x^3}} \right)}}{{dx}} = b\dfrac{{d\left( {{x^3}} \right)}}{{dx}}\], \[\dfrac{{d\left( {c{x^2}} \right)}}{{dx}} = c\dfrac{{d\left( {{x^2}} \right)}}{{dx}}\], \[\dfrac{{d\left( {dx} \right)}}{{dx}} = d\dfrac{{d\left( x \right)}}{{dx}}\], and \[\dfrac{{d\left( e \right)}}{{dx}} = e\dfrac{{d\left( 1 \right)}}{{dx}}\] in the equation \[\dfrac{{dy}}{{dx}} = \dfrac{{d\left( {a{x^4}} \right)}}{{dx}} + \dfrac{{d\left( {b{x^3}} \right)}}{{dx}} + \dfrac{{d\left( {c{x^2}} \right)}}{{dx}} + \dfrac{{d\left( {dx} \right)}}{{dx}} + \dfrac{{d\left( e \right)}}{{dx}}\], we get
\[ \Rightarrow \dfrac{{dy}}{{dx}} = a\dfrac{{d\left( {{x^4}} \right)}}{{dx}} + b\dfrac{{d\left( {{x^3}} \right)}}{{dx}} + c\dfrac{{d\left( {{x^2}} \right)}}{{dx}} + d\dfrac{{d\left( x \right)}}{{dx}} + e\dfrac{{d\left( 1 \right)}}{{dx}}\]
The derivative of the function of the form \[f\left( x \right) = {x^n}\] is written as \[f'\left( x \right) = n{x^{n - 1}}\].
The derivative of a constant is always 0.
Therefore, the equation becomes
$ \Rightarrow \dfrac{{dy}}{{dx}} = a\left( {4{x^3}} \right) + b\left( {3{x^2}} \right) + c\left( {2x} \right) + d\left( {1 \times {x^0}} \right) + e\left( 0 \right) \\
   \Rightarrow \dfrac{{dy}}{{dx}} = a\left( {4{x^3}} \right) + b\left( {3{x^2}} \right) + c\left( {2x} \right) + d\left( {1 \times 1} \right) + 0 \\
   \Rightarrow \dfrac{{dy}}{{dx}} = a\left( {4{x^3}} \right) + b\left( {3{x^2}} \right) + c\left( {2x} \right) + d\left( 1 \right) \\ $
Multiplying the terms of the expression, we get the first derivative as
\[ \Rightarrow \dfrac{{dy}}{{dx}} = 4a{x^3} + 3b{x^2} + 2cx + d\]
Now, we will differentiate the first derivative to find the second derivative.
Differentiating both sides of the equation with respect to \[x\], we get
$ \Rightarrow \dfrac{{{d^2}y}}{{d{x^2}}} = \dfrac{{d\left( {4a{x^3} + 3b{x^2} + 2cx + d} \right)}}{{dx}} \\
   \Rightarrow \dfrac{{{d^2}y}}{{d{x^2}}} = \dfrac{{d\left( {4a{x^3}} \right)}}{{dx}} + \dfrac{{d\left( {3b{x^2}} \right)}}{{dx}} + \dfrac{{d\left( {2cx} \right)}}{{dx}} + \dfrac{{d\left( d \right)}}{{dx}} \\ $
Simplifying the expression, we get
\[ \Rightarrow \dfrac{{{d^2}y}}{{d{x^2}}} = 4a\dfrac{{d\left( {{x^3}} \right)}}{{dx}} + 3b\dfrac{{d\left( {{x^2}} \right)}}{{dx}} + 2c\dfrac{{d\left( x \right)}}{{dx}} + d\dfrac{{d\left( 1 \right)}}{{dx}}\]
Therefore, we get
\[ \Rightarrow \dfrac{{{d^2}y}}{{d{x^2}}} = 4a\left( {3{x^2}} \right) + 3b\left( {2x} \right) + 2c\left( 1 \right) + d\left( 0 \right)\]
Multiplying the terms of the expression, we get the first derivative as
\[ \Rightarrow \dfrac{{{d^2}y}}{{d{x^2}}} = 12a{x^2} + 6bx + 2c\]
This is the second derivative of \[y = a{x^4} + b{x^3} + c{x^2} + dx + e\].
Now, a function \[f\left( x \right)\] has points of inflection if its second derivative is equal to zero, that is \[f''\left( x \right) = 0\].
Therefore, the condition such that \[y = a{x^4} + b{x^3} + c{x^2} + dx + e\] has points of inflection is
\[ \Rightarrow \dfrac{{{d^2}y}}{{d{x^2}}} = 12a{x^2} + 6bx + 2c = 0\]
We will find the roots of the quadratic equation \[12a{x^2} + 6bx + 2c = 0\] using the quadratic formula to find the required conditions.
The quadratic formula states that the roots of a quadratic equation \[A{x^2} + Bx + C = 0\] are given by \[x = \dfrac{{ - B \pm \sqrt D }}{{2A}}\], where \[D\] is the discriminant given by the formula \[D = {B^2} - 4AC\].
First, let us find the value of the discriminant.
Comparing the equation \[12a{x^2} + 6bx + 2c = 0\] with the standard form of a quadratic equation \[A{x^2} + Bx + C = 0\], we get
\[A = 12a\], \[B = 6b\], and \[C = 2c\]
Substituting \[A = 12a\], \[B = 6b\], and \[C = 2c\] in the formula for discriminant, we get
\[ \Rightarrow D = {\left( {6b} \right)^2} - 4\left( {12a} \right)\left( {2c} \right)\]
Simplifying the expression, we get
\[ \Rightarrow D = 36{b^2} - 96ac\]
Now, substituting \[A = 12a\], \[B = 6b\], and \[D = 36{b^2} - 96ac\] in the quadratic formula, we get
\[ \Rightarrow x = \dfrac{{ - 6b \pm \sqrt {36{b^2} - 96ac} }}{{2 \times 12a}}\]
Therefore, we get
\[\Rightarrow \sqrt {36{b^2} - 96ac} > 0\]
Squaring both sides, we get
\[ \Rightarrow 36{b^2} - 96ac > 0\]
Dividing both sides by 12, we get
$\Rightarrow \dfrac{{36{b^2} - 96ac}}{{12}} > \dfrac{0}{{12}} \\
\Rightarrow 3{b^2} - 8ac > 0 \\ $

Therefore, the condition for \[y = a{x^4} + b{x^3} + c{x^2} + dx + e\] to have points of inflection is \[3{b^2} - 8ac > 0\].

Note:
We simplified \[\dfrac{{dy}}{{dx}} = \dfrac{{d\left( {a{x^4} + b{x^3} + c{x^2} + dx + e} \right)}}{{dx}}\] to \[\dfrac{{dy}}{{dx}} = \dfrac{{d\left( {a{x^4}} \right)}}{{dx}} + \dfrac{{d\left( {b{x^3}} \right)}}{{dx}} + \dfrac{{d\left( {c{x^2}} \right)}}{{dx}} + \dfrac{{d\left( {dx} \right)}}{{dx}} + \dfrac{{d\left( e \right)}}{{dx}}\] and \[\dfrac{{{d^2}y}}{{d{x^2}}} = \dfrac{{d\left( {4a{x^3} + 3b{x^2} + 2cx + d} \right)}}{{dx}}\] to \[\dfrac{{{d^2}y}}{{d{x^2}}} = \dfrac{{d\left( {4a{x^3}} \right)}}{{dx}} + \dfrac{{d\left( {3b{x^2}} \right)}}{{dx}} + \dfrac{{d\left( {2cx} \right)}}{{dx}} + \dfrac{{d\left( d \right)}}{{dx}}\]. This is because the right hand side is a derivative of the sum of functions. The derivative of the sum of two functions is the sum of the derivatives of the two functions, that is \[\dfrac{{d\left[ {f\left( x \right) + g\left( x \right)} \right]}}{{dx}} = \dfrac{{d\left[ {f\left( x \right)} \right]}}{{dx}} + \dfrac{{d\left[ {g\left( x \right)} \right]}}{{dx}}\].