Question

What are the minimum and maximum values of the function $x^5-5x^4+5x^3-10$.

Answer 1

Step 1: Finding the critical points

Given: $f\left(x\right)=x^5-5x^4+5x^3-10$

Differentiate the given function with respect to $x$.

$f\text{'}\left(x\right)=5x^4-20x^3+15x^2$ $.......\left(1\right)$

For critical point substitute $f\text{'}\left(x\right)=0$. We get

$5x^4-20x^3+15x^2=0$

$\Rightarrow$ $5x^2\left(x^2-4x+3\right)=0$

$\Rightarrow$ $5x^2\left(x^2-3x-x+3\right)=0$

$\Rightarrow$ $5x^2\left[x\left(x-3\right)-1\left(x-3\right)\right]=0$

$\Rightarrow$ $5x^2\left(x-3\right)\left(x-1\right)=0$

$\Rightarrow$$5x^2=0,x-3=0,x-1=0$

$\Rightarrow$$x=0,x=3,x=1$

Thus,$x=0,1,3$ these are the critical points.

Step 2: Substituting the value of critical points in the second derivative of the function

Differentiate equation $\left(1\right)$ with respect to $x$, we get

$f\text{'}\text{'}\left(x\right)=20x^3-60x^2+30x$

Substitute $x=0$, we get

$f\text{'}\text{'}\left(0\right)=20\left(0\right)-60\left(0\right)+30\left(0\right)$

$\Rightarrow$$f\text{'}\text{'}\left(0\right)=0$

There is no maximum and no minimum value at $x=0$.

Substitute $x=1$, we get

$f\text{'}\text{'}\left(1\right)=20{\left(1\right)}^3-60{\left(1\right)}^2+30\left(1\right)$

$\Rightarrow$ $f\text{'}\text{'}\left(1\right)=20\left(1\right)-60\left(1\right)+30\left(1\right)$

$\Rightarrow$ $f\text{'}\text{'}\left(1\right)=20-60+30$

$\Rightarrow$ $f\text{'}\text{'}\left(1\right)=-10<0$

The function $f\left(x\right)$ is maximum at $x=1$.

Substitute $x=3$, we get

$f\text{'}\text{'}\left(3\right)=20{\left(3\right)}^3-60{\left(3\right)}^2+30\left(3\right)$

$\Rightarrow$ $f\text{'}\text{'}\left(3\right)=20\left(27\right)-60\left(9\right)+90$

$\Rightarrow$ $f\text{'}\text{'}\left(3\right)=540-540+90$

$\Rightarrow$ $f\text{'}\text{'}\left(3\right)=90>0$

The function $f\left(x\right)$ is minimum at $x=3$.

Step 3: Finding the minimum and maximum values of the function

The maximum value of $f\left(x\right)$ is $f\left(1\right)$.

$f\left(1\right)={\left(1\right)}^5-5{\left(1\right)}^4+5{\left(1\right)}^3-10$

$\Rightarrow$ $f\left(1\right)=1-5+5-10$

$\Rightarrow$ $f\left(1\right)=-9$

The minimum value of $f\left(x\right)$ is $f\left(3\right)$.

$f\left(3\right)={\left(3\right)}^5-5{\left(3\right)}^4+5{\left(3\right)}^3-10$

$\Rightarrow$ $f\left(3\right)=243-5\left(81\right)+5\left(27\right)-10$

$\Rightarrow$ $f\left(3\right)=243-405+135-10$

$\Rightarrow$ $f\left(3\right)=-37$

Hence, for the function $x^5-5x^4+5x^3-10$ the minimum value is $-37$ and the maximum value is $-9$.