Find the points of local maxima or local minima, functions, using the first derivative test. Also, find the local maximum or local minimum values, as the case may be f (x) = x3 (x – 1)2

Asked by Sakshi | 1 year ago |  81

##### Solution :-

Given, f(x) = x3(x – 1)2

Differentiate with respect to x, we get,

f ‘(x) = 3x2(x – 1)2 + 2x3(x – 1)

= (x – 1) (3x2(x – 1) + 2x3)

= (x – 1) (3x3 – 3x2 + 2x3)

= (x – 1) (5x3 – 3x2)

= x2 (x – 1) (5x – 3)

For all maxima and minima,

f ’(x) = 0

= x2(x – 1) (5x – 3) = 0

By solving the above equation we get

x =0, 1, $$\dfrac{3}{5}$$

At x =$$\dfrac{3}{5}$$, f’(x) changes from negative to positive

Since, x = $$\dfrac{3}{5}$$ is a point of Minima

At x =1, f‘(x) changes from positive to negative

Since, x =1 is point of maxima.

Answered by Sakshi | 1 year ago

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