"Absolute maxima and minima Determine the location and..." was answered by our community

druczekq4

Answered question

2021-12-06

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 \sqrt{2 - x^{2}} o n [- \sqrt{2}, \sqrt{2}]

Himin1945

Beginner2021-12-07Added 12 answers

Step 1
Given function is:

f (x) = x \cdot \sqrt{2 - x^{2}}

Differentiating with respect to x we get,

f^{'} (x) = \sqrt{2 - x^{2}} + x \cdot \frac{- 2 x}{2 \cdot \sqrt{2 - x^{2}}}

f^{'} (x) = \sqrt{2 - x^{2}} - \frac{x^{2}}{\sqrt{2 - x^{2}}}

f^{'} (x) = \frac{2 - x^{2} - x^{2}}{\sqrt{2 - x^{2}}}

f^{'} (x) = 2 \frac{1 - x^{2}}{\sqrt{2 - x^{2}}}

Step 2
To get the point of extremum we have:
f'(x) = 0

2 \frac{1 - x^{2}}{\sqrt{2 - x^{2}}} = 0

1 - x^{2} = 0

(1+x)(1-x)=0
x=-1; x=1
When x = 1, we get:

f (x) = 1 \cdot \sqrt{2 - 1^{2}} = 1

When x = -1, we get:

f (x) = (- 1) \cdot \sqrt{2 - {(- 1)}^{2}} = - 1

Therefore the points of extremum are: (1, 1) and (-1, -1).

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