Question

How do you find the maximum, minimum and inflection points and concavity for the function $y = {x^4} - 8{x^3}$

Answer 1

Hint: We need to take derivatives of the function two times to get the maximum or minimum points. We first equate the first derivative to $0$ and find the points. If at a point the second derivative gives a value strictly greater than $0$ then the point is a minimum point and if it is strictly less than $0$ then the point is a maximum point.
We know the derivative of $a{x^n}$.
$\therefore \dfrac{{d(a{x^n})}}{{dx}} = na{x^{(n - 1)}}$.
We calculate the zeroes of the second derivative of the function to get the inflection points. The inflection point is the point in the graph where we see that the concavity changes. If a graph gives a positive value then the function is convex in that interval and concave if the value is negative.

Complete step-by-step answer:
First, we start by finding the minimum and maximum points of the function $y = {x^4} - 8{x^3}$.
We take the first derivative of the given function. This only requires the power rule.
$
\therefore \dfrac{{dy}}{{dx}} = \dfrac{d}{{dx}}({x^4} - 8{x^3}) \\
\Rightarrow \dfrac{{dy}}{{dx}} = 4{x^3} - 8 \times 3{x^2} \\
\Rightarrow \dfrac{{dy}}{{dx}} = 4{x^3} - 24{x^2} \\
$
The derivative is the instantaneous slope of a line, and one needs to find the points when the derivative is $0$.
This will help us to determine the maximum and minimum of the original function. Since there is no constant term in this derivative, we can solve this by simply factorization. Then we find the factors by using the zero product principle.
$
\dfrac{{dy}}{{dx}} = 0 \\
\Rightarrow 4{x^3} - 24{x^2} = 0 \\
\Rightarrow 4{x^2}(x - 6) = 0 \\
\therefore x = 0,x = 6 \\
$
Putting these values in the original function we get the two points as:
When $x = 0$,
$
y = {(0)^4} - 8{(0)^3} \\
\Rightarrow y = 0 \\
$
And when $x = 6$,
$
y = {\left( 6 \right)^4} - 8{(6)^3} \\
\Rightarrow y = {6^3}(6 - 8) \\
\Rightarrow y = {6^3} \times ( - 2) \\
\Rightarrow y = 216 \times ( - 2) \\
\Rightarrow y = - 432 \\
$
Therefore the required points are $(0,0)$ and $(6, - 432)$.
We find the second derivative to check if it is a maximum or minimum point.
$
\dfrac{{{d^2}y}}{{d{x^2}}} = \dfrac{d}{{dx}}\left( {\dfrac{{dy}}{{dx}}} \right) \\
\Rightarrow \dfrac{{{d^2}y}}{{d{x^2}}} = \dfrac{d}{{dx}}\left( {4{x^3} - 24{x^2}} \right) \\
\Rightarrow \dfrac{{{d^2}y}}{{d{x^2}}} = 4 \times 3{x^2} - 24 \times 2x \\
\Rightarrow \dfrac{{{d^2}y}}{{d{x^2}}} = 12{x^2} - 48x \\
$
At $x = 6$, we have the value of the second derivative as:
$
= 12{x^2} - 48x \\
= 12{(6)^2} - 48(6) \\
= 144 \\
$
As the second derivative is greater than $0$ so the point $(6, - 432)$ is a point of minimum.
To find the inflection point we calculate the zeroes of the second derivative. The zeroes of the second derivative are as follows:
$
\dfrac{{{d^2}y}}{{d{x^2}}} = 0 \\
\Rightarrow 12{x^2} - 48x = 0 \\
\Rightarrow 12x(x - 4) = 0 \\
\therefore x = 0,x = 4 \\
$
We divide the interval into three ranges $( - \infty ,0),(0,4),(4,\infty )$.
When $x \in ( - \infty ,0)$, $y > 0$.
When $x \in (0,4),y < 0$.
Lastly when $x \in (4,\infty ),y > 0$.
When $x = 4,y = {(4)^4} - 8{(4)^3} = - 256$
The inflection points are $(0,0)$ and $(4, - 256)$
The sign of this graph will give us the concavity of the original graph
The curve is convex when $x \in ( - \infty ,0) \cup (4,\infty )$
The curve is concave when $x \in \left( {0,4} \right)$

Note: The inflection can sometimes be a stationary point. A function may or may not have a definite minimum or maximum point always. We need to look carefully left and right of the inflection points to look for the concavity of the function.