Find the points at which the function f given by f(x)=(x-2)^4(x+1)^3 has: 1. local maxima. 2. local minima. 3. point of inflexion.

The given function is f(x)=(x-2)^4(x+1)^3. f'(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(7x-2) Now, f'(x)=0 x=- and x=2 7 or x=2 Now, for values of x close to 2 7 and to the left of 2 7 , f'(x) > 0. Also, for values of x close to 2 7 and to the right of 2 7 f'(x) 0. Thus, x = 2 is the point of local minima. Now, as the value of x varies through -1, f'(x) does not changes its sign. Thus, x = -1 is the point of inflexion.

Find the points at which the function f given by f(x)=(x-2)^4(x+1…

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Question

Find the points at which the function f given by $\text{f}\text{(x)}=\text{(x}-2)^4(\text{x}+1)^3$ has:

local maxima.
local minima.
point of inflexion.

✓

Answer

The given function is $\text{f}\text{(x)}=\text{(x}-2)^4(\text{x}+1)^3.$
$\therefore\ \text{f}'\text{(x)}=4\text{(x}-2)^3(\text{x}+1)^3+3(\text{x}+1)^2(\text{x}-2)^4$
$=(\text{x}-2)^3(\text{x}+1)^2[4(\text{x}+1)+3(\text{x}-2)]$
$=(\text{x}-2)^3(\text{x}+1)^2(7\text{x}-2)$
Now, $\text{f}'(\text{x})=0\Rightarrow\ \text{x}=-\text{ and x}=\frac{2}{7}\text{ or x}=2$
Now, for values of x close to $\frac{2}{7}$ and to the left of $\frac{2}{7},$ f'(x) > 0. Also, for values of x close to $\frac{2}{7}$ and to the right of $\frac{2}{7}$ f'(x) < 0.
Thus, $\text{x}=\frac{2}{7}$ is the point of local maxima.
Now, for values of x close to 2 and to the left of 2, f'(x) < 0. Also, for values of x close to 2 and to the right of 2, f'(x) > 0.
Thus, x = 2 is the point of local minima.
Now, as the value of x varies through -1, f'(x) does not changes its sign.
Thus, x = -1 is the point of inflexion.