Answer :

It is given that 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 = - 1 and x = or x = 2


Now, for values of x close to and to the left of


f’(x) > 0.


Also, for values of x close to and to the right of , f’(x) < 0.


Then, x = 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.


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


Now, as the value of x varies through - 1, f’(x) does not changes its sign.


Then, x = - 1 is the point of inflexion.


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