What-is-max-and-min-value-of-y-4-x-2-1-x-2-

Question Number 136138 by bramlexs22 last updated on 19/Mar/21

W h a t i s m a x a n d m i n v a l u e

o f y = 4 - x^{2} - \sqrt{1 - x^{2}} ?

Answered by ajfour last updated on 19/Mar/21

l e t 1 - x^{2} = s ⩾ 0

\Rightarrow

y = 3 + s - \sqrt{s}

\frac{d y}{d s} = 1 - \frac{1}{2 \sqrt{s}} = 0 \Rightarrow

s = \frac{1}{4}

y = 3 + \frac{1}{4} - \frac{1}{2} = \frac{11}{4}

\frac{d^{2} y}{{d s}^{2}} ∣_{s = \frac{1}{4}} = (\frac{1}{4 s \sqrt{s}}) ∣_{s = \frac{1}{4}} > 0

h e n c e y_{m i n} = \frac{11}{4}

a n d f o r s = 0 (x = \pm 1)

y = 3

f o r s = 1 (x = 0)

y = 3

h e n c e y_{m a x} = 3

Answered by mr W last updated on 19/Mar/21

y = 3 + (1 - x^{2}) - \sqrt{1 - x^{2}}

y = \frac{11}{4} + {(\sqrt{1 - x^{2}} - \frac{1}{2})}^{2} ⩾ \frac{11}{4}

y_{m i n} = \frac{11}{4} w h e n 1 - x^{2} = \frac{1}{4}, x \pm \frac{\sqrt{3}}{2}

y_{m a x} = \frac{11}{4} + {(\pm \frac{1}{2})}^{2} = 3 w h e n x = 0 o r \pm 1

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