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Application of Derivatives

Name: Zigya - For The Curious Learner
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Long Answer Type

251. Find the local maxima or local minima, if any, of following functions using the first derivative test only. Find also the local maximum and the local minimum values, as the case may be:

straight x cubed left parenthesis straight x minus 1 right parenthesis squared

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252. Find the local maxima or local minima, if any, of following functions using the first derivative test only. Find also the local maximum and the local minimum values, as the case may be:

straight x cubed left parenthesis 2 straight x minus 1 right parenthesis cubed

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253. Find the local maxima or local minima, if any, of following functions using the first derivative test only. Find also the local maximum and the local minimum values, as the case may be:
$negative left parenthesis straight x minus 1 right parenthesis cubed space left parenthesis straight x plus 1 right parenthesis squared$

Let f (x) = – (x – 1)³ ( x + 1 )²∴ f ' (x) = – [(x – 1)³. 2(x + 1) + (x + 1)². 3(x – 1)²]
= – (x – 1)²(x + 1) [2 (x – 1) + 3 (x + 1)] = – (x – 1)² (x + 1) (5 x + 1)
f ' (x) = 0 ⇒ – (x – 1)²(x + 1) (5 x + 1) = 0
$rightwards double arrow space space straight x space equals 1 comma space space space 1 comma space space space minus 1 fifth$ .
When x < 1 slightly, f ' (x) = – [( + )( + )( + )] = – ve
When x > 1 slightly, f ' (x) = – [( + )( + )( + )] = – ve
∴ at x = 1 , f '(x) does not change sign ,
∴ x = 1 is a point of inflexion.
When x < – 1 slightly. f ' (x) = – [( + )( – )( – )] = – ve
When x > – 1 slightly. f ' (x) = – [( + )( + ) ( – )] = + ve
∴ at x = – 1, f ' (x) changes from – ve to + ve
∴ f (x) has local minimum value at x = – 1 and this local minimum value = 0.
$when space straight x less than negative 1 fifth space slightly comma space space space straight f space apostrophe left parenthesis straight x right parenthesis space equals space minus open square brackets left parenthesis plus right parenthesis thin space left parenthesis plus right parenthesis thin space left parenthesis negative right parenthesis close square brackets space space equals space plus ve When space straight x space greater than 1 fifth space slightly comma space space space straight f space apostrophe left parenthesis straight x right parenthesis space equals space left square bracket left parenthesis plus right parenthesis thin space left parenthesis plus right parenthesis thin space left parenthesis plus right parenthesis right square bracket space equals space minus ve therefore space space space at space straight x space equals space minus 1 fifth comma space space space space space straight f space apostrophe left parenthesis straight x right parenthesis space changes space from space plus ve space to space minus ve therefore space space space space straight f left parenthesis straight x right parenthesis space has space local space maximum space value space at space straight x space equals space minus 1 fifth and space this space value space space equals space minus open parentheses 6 over 5 close parentheses cubed space space open parentheses 4 over 5 close parentheses cubed space equals space fraction numerator 6 cubed cross times 4 squared over denominator 5 to the power of 5 end fraction space equals space 3456 over 3125. space$

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Short Answer Type

254.

Find all the local maxima or minima of the function
f (x) = x³– 12x.

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255. Find all the points of local maxima and local minima of the function f given by f (x) = x³ – 27 x + 3. Also, find local maximum and local minimum value of f.

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