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edited | 113 views
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Hi,

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minima and maxima or minimum value and maximum value?
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Maxima at point x = 0.

Minima at point x = -3, 3.

+1 vote

$f(x)=\sqrt{36-4x^{2}}$   here  $-3\leq x\leq3$

$f(x)=(36-4x^{2})^{\frac{1}{2}}$------------>$(1)$

differentiate both sides with respect to $'x'$

$f'(x)=\frac{1}{2}(36-4x^{2})^{(\frac{1}{2}-1)}.(0-4.2x)$

$f'(x)=\frac{1}{2}(36-4x^{2})^{\frac{-1}{2}}.(-8x)$

$f'(x)=\frac{1}{2}[\frac{1}{(36-4x^{2})^{\frac{1}{2}}}].(-8x)$

$f'(x)=\frac{-4x}{(36-4x^{2})^{\frac{1}{2}}}$

$f'(x)=\frac{-4x}{2.(9-x^{2})^{\frac{1}{2}}}$

$f'(x)=\frac{-2x}{(9-x^{2})^{\frac{1}{2}}}$------------->$(2)$

for statinary points $f'(x)=0$

$\Rightarrow \frac{-2x}{(9-x^{2})^{\frac{1}{2}}}=0$

$\Rightarrow x=0$

$(or)$

$(9 - x^{2})^{-1/2} = 0.$

we can write it like

$0^{-2} = ( (9- x^{2})^{-1/2})^{-2}$

$9 - x^{2} = 0$

$x = -3,3$

So, we got three stationary points or critical points $x=0,-3,3$

now,$f(-3)=\sqrt{(36-4(-3)^{2})}=\sqrt{(36-36)}=0$

$f(0)=\sqrt{(36-4(0)^{2})}=\sqrt{36}=6$

$f(3)=\sqrt{(36-4(3)^{2})}=\sqrt{(36-36)}=0$

Absolute minimum $f(x)=0$ at $x=-3,3$

Absolute maximum  $f(x)=6$ at $x=0$

edited
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-3 is wrong.. rest values are correct..i'm not sure whether by absolute minima and minima they're referring to global minima and maxima?
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Why $-3$ is wrong?
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Local minima and local maxima is the minimum and maximum of a function in a particular region while absolute maxima and absolute minima is the maximum and minimum value of the overall function.

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then what is global minima and global maxima?
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in the answer it's given 0 and 3 only
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ok.... thank u @Lakshman Patel RJIT

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:)
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