Find the points at which the function

The given function is $f(x)=(x-2)^{4}(x+1)^{3}$.

$\begin{aligned} \therefore f^{\prime}(x) &=4(x-2)^{3}(x+1)^{3}+3(x+1)^{2}(x-2)^{4} \\ &=(x-2)^{3}(x+1)^{2}[4(x+1)+3(x-2)] \\ &=(x-2)^{3}(x+1)^{2}(7 x-2) \end{aligned}$

Now, $f^{\prime}(x)=0 \Rightarrow x=-1$ and $x=\frac{2}{7}$ or $x=2$

Now, for values of $x$ close to $\frac{2}{7}$ and to the left of $\frac{2}{7}, f^{\prime}(x)>0$. Also, for values of $x$ close to $\frac{2}{7}$ and to the right of $\frac{2}{7}, f^{\prime}(x)<0$.

Thus, $x=\frac{2}{7}$ is the point of local maxima.

Now, for values of $x$ close to 2 and to the left of $2, f^{\prime}(x)<0$. Also, for values of $x$ close to 2 and to the right of $2, f^{\prime}(x)>0$.

Thus, x = 2 is the point of local minima.

Now, as the value of x varies through −1,does not changes its sign.

Thus, x = −1 is the point of inflexion.