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Example 18

If f (x) = x4 - 8x2, determine all local extrema (maxima and minima) for the function.

Solution : f (x) = x4 - 8x2
\ f '(x) = 4x3 - 16x

For critical points (i.e. roots of f ' (x) ) f ' (x) = 0 Þ

4x3 - 16x = 0 or

4x (x2 - 4) = 0 or

x = 0, x = -2, x = 2

Now for h > 0 (very small)

f ' (0 - h) = 4 (0 + h)3 - 16 (0 - h) = 4h3 + 16h > 0

and f'(0+h) = 4 (0 + h)3 - 16 (0 + h) = 4h3 + 24h2 + 32h >0

i.e.f'(x) changes its sign from negative to positive around x =-2
\ f ' (x) has a Local minimum at x = -2 i.e. (-2, -16)

Similarly we can show that f '(2 - h) < 0 and f ' (2 + h) > 0
\ f '(x) changes its sign from negative to positive around x = 2
\ f '(x) has a local minimum at x = 2
i.e. (2, -16) Your browser does not support the IFRAME tag.

Example 19

Find the maximum and minimum values of f (x) = Solution : f (x) = \ f ' (x) = \ f '(x) = = 3 (x - 5) (x - 1)

For critical points of f (x), f ' (x) = 0 Þ x = 5 and x = 1

Now 1 - nbd of 5 we have

f ' (x) < 0 when x < 5

f ' (x) > 0 when x > 1

\ f (5) is a minimum value and f(5) = - 22

\ f (1) is maximum value and f(1) = 10 Next in 1 - nbd of 1, we have

f ' (x) > 0 when x < 1

f ' (x) < 0 when x > 1

\ f (1) is maximum value and f (1) = 10 Index

5.1 Tangent And Normal Lines
5.2 Angle Between Two Curves
5.3 Interpretation Of The Sign Of The Derivative
5.4 Locality Increasing Or Decreasing Functions 5.5 Critical Points
5.6 Turning Points
5.7 Extreme Value Theorem
5.8 The Mean-value Theorem
5.9 First Derivative Test For Local Extrema
5.10 Second Derivative Test For Local Extrema
5.11 Stationary Points
5.12 Concavity And Points Of Inflection
5.13 Rate Measure (distance, Velocity And Acceleration)
5.14 Related Rates
5.15 Differentials : Errors And Approximation All Contents Copyright © All rights reserved.
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