Question

Find the points at which the function f given by $f\left(x\right)=(x-2{)}^4(x+1{)}^3$ has
(a) Local maxima
(b) Local minima
(c) Point of inflection

Answer 1

The given equation is $f\left(x\right)=(x-2{)}^4(x+1{)}^3$
On differentiating w.r.t.x, we get
$f^{\prime }\left(x\right)=(x-2{)}^{4}3(x+1{)}^2+(x+1{)}^{3}4(x-2{)}^3$
$=(x-2{)}^3(x+1{)}^2\left\{3\right(x-2)+4(x+1\left)\right\}=(x-2{)}^3(x+1{)}^2\{7x-2\}$
Clearly, f'(x) exists at every point of the domain.
For maximum or minimum f'(x) = 0
$\Rightarrow (x-2{)}^3(x+1{)}^2(7x-2)=0$
$\Rightarrow x=2,-1,\frac{2}{7}$
At $x=2,$
$f^{\prime }\left(x\right)$ & $(x-2{)}^3$ & $(x+1{)}^2$ & $(7x-2)$ & Result
\hline
$x$slightly $<2$ & -ve & +ve & +ve & -ve
$x$slightly $>2$ & +ve & +ve & +ve & -ve
$\therefore f^{\prime }\left(x\right)$ change sign from -ve to +ve. Thus, x = 2 is s point of local minima.
At$x=1,$
f'(x) & $(x-2{)}^3$ & $(x+1{)}^2$ & $(7x-2)$ & Result
x slightly < -1 & -ve & +ve & -ve & +ve
x slightly > -1 & -ve & +ve & -ve & +ve
$\therefore f^{\prime }\left(x\right)$does not change the sign. Thus, $x=-1$ is a point of inflection.
At$x=\frac{2}{7}$,
$(x-2{)}^3$ & $(x+1{)}^2$ & $(7x-2)$ & Result
x slightly $<\frac{2}{7}$ & -ve & +ve & -ve & +ve
x slightly $>\frac{2}{7}$ & -ve & +ve & +ve & +ve
$\therefore f^{\prime }\left(x\right)$ changes sign from +ve to -ve. Thus, $x=\frac{2}{7}$ is a point of local maxima.
So, we can say as the value of x varies through -1, f'(x) does not change its sign. Thus, x = - 1 is the point of inflection.
Note If the derivative of a function does not change the sign, then it has a point of inflection at that point