+1 vote
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asked in Calculus | 133 views

Absolute Minimum will be at $x = 3$ and $x = -3$.

Absolute minimum of $f\left ( x \right )$ is the minimum possible value that $f\left ( x \right )$ can ever attain.

Since a square root never spits out a $-ve$ value, the minimum value that $f\left ( x \right )$ can attain is $0$.

Now to make $f\left ( x \right ) = 0$, $\left ( 36 -4x^2 \right )$ must be equal to $0$.

On solving $\left ( 36 -4x^2 \right )= 0$, we get $x = 3$ & $x = -3$.

So at $x = 3$ & $x = -3$ the function $f\left ( x \right )$ will be at its absolute minimum.

that is, $f\left ( 3 \right ) = f\left ( -3 \right ) = 0$.

Also the domain of $f\left ( x \right )$ is $\left [ -3, 3 \right ]$ & the range is $\left [ 0, 6 \right ]$

selected
@ Anurag : why √36 is not considered as  +6 & -6 both ?

Amsar, sorry I don't know why $\sqrt{36}$ is not considered as $+6$ & $-6$,
but $+\sqrt{36}$ is considered as $+6$ & $-\sqrt{36}$ is considered as $-6$.

ok.. :)
+1 vote

f(x)=√(36-4x2)

=2√(9-x2)

For getting absolute minimum f(x) should be 0

So, 9-x2 =0

x=3,-3