"Absolute maxima and minima Determine the location and..." solved and explained

Concepcion Hale

Answered question

2021-12-12

Absolute maxima and minima Determine the location and value of the absolute extreme value of ƒ on the given interval, if they exist.

f (x) = x^{\frac{1}{3}} (x + 4) o n [- 27, 27]

Philip Williams

Beginner2021-12-13Added 39 answers

Step 1
Absolute maxima and minima:

f (x) = x^{\frac{1}{3}} (x + 4), [- 27, 27]

f^{'} (x) = x^{\frac{1}{3}} \cdot \frac{d}{d x} (x + 4) + (x + 4) \cdot \frac{d}{d x} (x^{\frac{1}{3}})

f^{'} (x) = x^{\frac{1}{3}} + (x + 4) \cdot \frac{1}{3} \cdot x^{\frac{- 2}{3}}

f^{″} (x) = \frac{1}{3} x^{\frac{- 2}{3}} + \frac{x + 4}{3} \cdot (\frac{- 2}{3}) \cdot x^{\frac{- 5}{3}}

= \frac{1}{3} x^{\frac{- 2}{3}} - \frac{2 (x + 4)}{9} \cdot x^{- \frac{5}{3}} > 0

Step 2
To find maxima or minima,
f'(x)=0

x^{\frac{1}{3}} + \frac{x + 4}{3} \cdot x^{\frac{- 2}{3}} = 0

3 x^{\frac{1}{3}} + x^{\frac{1}{3}} + 4 x^{\frac{- 2}{3}} = 0

4 x^{\frac{1}{3}} = - 4 x^{\frac{- 2}{3}}

x^{\frac{1}{3}} = - x^{\frac{- 2}{3}}

x=-1
f(x)=-3
The absolute minima is (-1,-3).

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