Question

How do you find all critical point and determine the min, max and inflection given $f\left( x \right)={{x}^{4}}-4{{x}^{3}}+20$ ?

Answer 1

First to find the critical point find the derivative of the function $f\left( x \right)$ i.e., ${f}'\left( x \right)=0$ . After evaluation find the values of $x$ which will then be the critical points of the function. Now find the second derivative and if it is a positive value then the function’s minimum is at the critical point. To find the inflection points, perform ${f}''\left( x \right)=0$ The values of $x$ hence obtained will be inflection points.

Complete step-by-step solution:
The given polynomial is $f\left( x \right)={{x}^{4}}-4{{x}^{3}}+20$
Now let us first find the first derivative of the function, ${f}'\left( x \right)$
$\Rightarrow {f}'\left( x \right)=4{{x}^{3}}-12{{x}^{2}}$
Now let us find the second derivative, ${f}''\left( x \right)$
$\Rightarrow {f}''\left( x \right)=12{{x}^{2}}-24{{x}^{{}}}$
To find the critical point we need to find ${f}'\left( x \right)=0$
$\Rightarrow {f}'\left( x \right)=4{{x}^{3}}-12{{x}^{2}}=0$
$\Rightarrow {{x}^{2}}\left( 4x-12 \right)=0$
Now split the above expression to get the factors.
$\Rightarrow x=0;\left( 4x-12=0 \right)$
$\Rightarrow x=0;x=\dfrac{12}{4}$
Hence the critical points are given by the values $x=0;x=3$
Now substitute these values in the function.
$f\left( 0 \right)={{0}^{4}}-4{{\left( 0 \right)}^{3}}+20=20$
$f\left( 3 \right)={{3}^{4}}-4{{\left( 3 \right)}^{3}}+20=-7$
The value of ${f}'\left( x \right)$ is decreasing from $\left[ -\infty ,0 \right)$ and further decreasing till $\left[ 0,3 \right)$.
Further, the curve increases from $\left[ 0,3 \right)\to \left[ 3,\infty \right)$
Hence the local minima will be $\left( 3,-7 \right)$
Now find the inflection points.
For that, we need to find ${f}''\left( x \right)=0$
$\Rightarrow {f}''\left( x \right)=12{{x}^{2}}-24x=0$
$\Rightarrow x\left( 12x-24 \right)=0$
Now split the above expression to get the factors of $x$ .
$\Rightarrow x=0;\left( 12x-24=0 \right)$
$\Rightarrow x=0;x=\dfrac{24}{12}$
Hence the values of inflection points are $x=0;x=2$
If the sign of ${f}'\left( x \right)$ does not change as $x$ increases through $c$ , such point is known as point of inflection.
The value of ${f}'\left( x \right)$ is decreasing from $\left[ -\infty ,0 \right)$ and further decreasing till $\left[ 0,3 \right)$.
Further, the curve decreases from $\left[ 0,2 \right)\to \left[ 2,3 \right)$
The sign of ${f}'\left( x \right)$ remains the same.
Now substitute these values to get the coordinates.
$f\left( 0 \right)={{0}^{4}}-4{{\left( 0 \right)}^{3}}+20=20$
$f\left( 2 \right)={{2}^{4}}-4{{\left( 2 \right)}^{3}}+20=4$
The inflection points are hence, $\left( 0,20 \right);\left( 2,4 \right)$

Note: If the sign of ${f}'\left( x \right)$ shifts from positive to negative as the value of $x$ increases through $c$ , ${f}'\left( x \right)>0$ at every point to the left of $c$ and ${f}'\left( x \right)<0$ to every point right of $c$ , then the point $c$ is said to be the local maxima.
If the sign of ${f}'\left( x \right)$ shifts from negative to positive as the value of $x$ increases through $c$ , ${f}'\left( x \right)>0$ at every point to the right of $c$ and ${f}'\left( x \right)<0$ to every point left of $c$ , then the point $c$ is said to be the local minima.