Find the points at which the function f given by $f\left(x\right) = \left(x-2\right)^4 \left(x+1\right)^3$ has local maxima, local minima and point of inflection.
$f\left(x\right) = \left(x-2\right)^4 \left(x+1\right)^3$
$\begin{aligned}
\therefore f'\left(x\right) & = 4 \left(x-2\right)^3 \left(x+1\right)^3 + 3 \left(x+1\right)^2 \left(x-2\right)^4 \\
& = \left(x-2\right)^3 \left(x+1\right)^2 \left[4\left(x+1\right)+3\left(x-2\right)\right] \\
& = \left(x-2\right)^3 \left(x+1\right)^2 \left(7x-2\right)
\end{aligned}$
For critical points, $f'\left(x\right) = 0$
i.e. $\left(x-2\right)^3 \left(x+1\right)^2 \left(7x-2\right) = 0$
$\implies$ $x-2=0$ or $x+1=0$ or $7x-2=0$
i.e. $x=2$ or $x=-1$ or $x = \dfrac{2}{7}$
Consider $x=2$
When $x<2$, let $x=1.9$
$\begin{aligned}
\text{Then, } f'\left(x\right) = f'\left(1.9\right) & = \left(1.9-2\right)^3 \left(1.9+1\right)^2 \left[\left(7\times 1.9\right) -2\right] \\
& = \left(-ve\right) \left(+ve\right) \left(+ve\right) \\
& = -ve
\end{aligned}$
When $x>2$, let $x=2.1$
$\begin{aligned}
\text{Then, } f'\left(x\right) = f'\left(2.1\right) & = \left(2.1-2\right)^3 \left(2.1+1\right)^2 \left[\left(7\times 2.1\right) -2\right] \\
& = \left(+ve\right) \left(+ve\right) \left(+ve\right) \\
& = +ve
\end{aligned}$
i.e. $f'\left(x\right)$ changes sign from negative to positive as x increases through 2.
$\implies$ $x=2$ is a point of local minimum.
Consider $x=-1$
When $x<-1$, let $x=-1.1$
$\begin{aligned}
\text{Then, } f'\left(x\right) = f'\left(-1.1\right) & = \left(-1.1-2\right)^3 \left(-1.1+1\right)^2 \left[7 \times\left(-1.1\right) -2\right] \\
& = \left(-ve\right) \left(+ve\right) \left(-ve\right) \\
& = +ve
\end{aligned}$
When $x>-1$, let $x=-0.9$
$\begin{aligned}
\text{Then, } f'\left(x\right) = f'\left(-0.9\right) & = \left(-0.9-2\right)^3 \left(-0.9+1\right)^2 \left[7 \times \left(-0.9\right) -2\right] \\
& = \left(-ve\right) \left(+ve\right) \left(-ve\right) \\
& = +ve
\end{aligned}$
i.e. $f'\left(x\right)$ does not change sign as x increases through $-1$.
$\implies$ $x=-1$ is a point of inflexion.
Consider $x = \dfrac{2}{7}$
When $x < \dfrac{2}{7}$, let $x = \dfrac{1.9}{7}$
$\begin{aligned}
\text{Then, } f'\left(x\right) = f'\left(\dfrac{1.9}{7}\right) & = \left(\dfrac{1.9}{7}-2\right)^3 \left(\dfrac{1.9}{7}+1\right)^2 \left[\left(7\times \dfrac{1.9}{7}\right) -2\right] \\
& = \left(-ve\right) \left(+ve\right) \left(-ve\right) \\
& = +ve
\end{aligned}$
When $x>\dfrac{2}{7}$, let $x=\dfrac{2.1}{7}$
$\begin{aligned}
\text{Then, } f'\left(x\right) = f'\left(\dfrac{2.1}{7}\right) & = \left(\dfrac{2.1}{7}-2\right)^3 \left(\dfrac{2.1}{7}+1\right)^2 \left[\left(7\times \dfrac{2.1}{7}\right) -2\right] \\
& = \left(-ve\right) \left(+ve\right) \left(+ve\right) \\
& = -ve
\end{aligned}$
i.e. $f'\left(x\right)$ changes sign from positive to negative as x increases through $\dfrac{2}{7}$.
$\implies$ $x=\dfrac{2}{7}$ is a point of local maximum.