# Find the points at which the function $$f$$ given by $$f (x) = (x – 2)^4 (x + 1)^3$$ has a (i) local maxima, (ii) local minima and the (iii) point of inflexion.

Toolbox:
• $\large\frac{d}{dx}$$(x^n)=nx^{n-1} • For maxima & minima f'(x)=0 Step 1: Given: Let f(x)=(x-2)^4.(x+1)^3 Differentiating with respect to x we get f'(x)=(x-2)^43(x+1)^2+(x+1)^3.4(x-2)^3 \qquad=(x-2)^3.(x+1)^2[3(x-2)+4(x+1)] \qquad=(x-2)^3.(x+1)^2(7x-2) \qquad=(x-2)^3.(x+1)^2[3(x-2)+4(x+1)] \qquad=7(x-2)^3(x+1)^2(x-\large\frac{2}{7}) For maxima and minima,f'(x)=0 \Rightarrow 7(x-2)^3(x+1)^2(x-\large\frac{2}{7})$$=0$
$\Rightarrow x=2,-1,\large\frac{2}{7}$
Step 2:
At $x=2$
When $x$ is slightly $<2$
$f'(x)=(-)(+)(+)=-ve$
When $x$ is slightly $>2$
$f'(x)=(+)(+)(+)=+ve$
$f'(x)$ changes the sign from -ve to +ve while crossing the point $x=2$
$\Rightarrow f(x)$ is minimum at $x=2$
Step 3:
At $x=-1$
When $x$ is slightly $<-1$
$f'(x)=(-)(+)(-)=+ve$
When $x$ is slightly $>-1$
$f'(x)=(-)(+)(-)=+ve$
$\Rightarrow f'(x)$ does not changes the sign while passing through -1
Hence x = -1 is a point of inflexion.
Step 4: