Question

find the extreme values of the function and where they occur. x^3-3x^2+1

Answer 1

u have to differentiate the function here

Answer 2

after differentiating u have to equate it to zero


\(\large\tt \begin{align} \color{black}{x^3-3x^2+1 \\~\\
\dfrac{d}{dx}(x^3)-\dfrac{d}{dx}(3x^2)+\dfrac{d}{dx}(1 )=0\\~\\
3x^2-6x+0=0\\~\\
3x^2-6x=0\\~\\~\\~\\
\Large \rm x=0\\~\\
or\\~\\
\Large \rm x=2
}\end{align}\)

Answer 3

so

when x=0
\(\large\tt \begin{align} \color{black}{y=x^3-3x^2+1 \\~\\

y=0^3-3\times 0^2+1 \\~\\
y=1

}\end{align}\)
when x=2
\(\large\tt \begin{align} \color{black}{y=x^3-3x^2+1 \\~\\

y=2^3-3\times 2^2+1 \\~\\
y=-3

}\end{align}\)

so the extreme points are

\(\Large (x,y)=(0,1)(max)\)

and


\(\Large (x,y)=(2,-3)(min)\)