Find the points at which the function f given by f (x) = (x - 2)4 (x + 1)3 has
(i) local maxima (ii) local minima (iii) point of inflexion
Solution:
The given function is f (x) = (x - 2)4 (x + 1)3
Thereofore,
On differentiating wrt x, we get
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, x = 2/7, x = 2
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 = 2/7 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 change its sign.
Thus, x = - 1 is the point of inflexion
NCERT Solutions Class 12 Maths - Chapter 6 Exercise ME Question 13
Find the points at which the function f given by f (x) = (x - 2)4 (x + 1)3 has (i) local maxima (ii) local minima (iii) point of inflexion
Summary:
f given by f (x) = (x - 2)4 (x + 1)3. x = 2/7 is the point of local maxima. x = 2 is the point of local minima. x = - 1 is the point of inflection
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