Given: fx=x 13x+12⇒f'x=3x 12x+12+2x 13x+1For nbsp;a nbsp;local nbsp;maximum nbsp;or nbsp;a nbsp;local nbsp;minimum, nbsp;we nbsp;must nbsp;h ...

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Rolle's Theorem

fx=x-13x+12

Question

f(x) = (x $-$ 1)³ (x+1)²

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Solution

$Given : f (x) = {(x - 1)}^{3} {(x + 1)}^{2} \Rightarrow f' (x) = 3 {(x - 1)}^{2} {(x + 1)}^{2} + 2 {(x - 1)}^{3} (x + 1) For a local maximum or a local minimum, we must have f' (x) = 0 \Rightarrow 3 {(x - 1)}^{2} {(x + 1)}^{2} + 2 {(x - 1)}^{3} (x + 1) = 0 \Rightarrow (x + 1) {(x - 1)}^{2} (3 x + 3 + 2 x - 2) = 0 \Rightarrow (x + 1) {(x - 1)}^{2} (5 x + 1) = 0 \Rightarrow x = - 1, 1 and \frac{- 1}{5}$

Since f '(x) changes from negative to positive when x increases through $\frac{- 1}{5}$ , x = $\frac{- 1}{5}$ is the point of local minima.
The local minimum value of f (x) at x = $\frac{- 1}{5}$ is given by
${(\frac{- 1}{5} - 1)}^{3} {(\frac{- 1}{5} + 1)}^{2} = \frac{- 3456}{3125}$

Since f '(x) changes from positive to negative when x increases through $- 1$ , x = $- 1$ is the point of local maxima.
The local maximum value of f (x) at x = $- 1$ is given by
${(- 1 - 1)}^{3} {(- 1 + 1)}^{2} = 0$

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