Find the absolute extrema of the function (if any exist) on each interval. (If an answer does not exist, enter DNE.) $$ f(x)=\sqrt{25-x^{2}} $$ (a) $$ [-5,5] $$ $$ \begin{array}{ll} \text { minimum (smaller } x \text {-value) } & (x, y)=(\square) \ \text { minimum (larger } x \text {-value) } & (x, y)=(\square) \ \text { maximum } & (x, y)=(\square) \end{array} $$ (b) $$ [-5,0) $$ minimum $$ (x, y)=(\square) $$ maximum $$ (x, y)=(\square) $$ (c) $$ (-5,5) $$ minimum $$ (x, y)=(\square) $$ maximum $$ (x, y)=(\square) $$ (d) $$ [1,5) $$ minimum $$ (x, y)=(\square) $$ maximum $$ (x, y)=(\square) $$

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(a) For $$ f(x)=\sqrt{25-x^2} $$ on $$[ -5,5 ]$$:

1. The endpoints are $$x=-5$$ and $$x=5$$. We have
   $$
   f(-5)=\sqrt{25-(-5)^2}=\sqrt{25-25}=0,\quad f(5)=0.
   $$
2. On the interior, the function is maximum when the radicand $$ 25-x^2 $$ is maximized. Since
   $$
   25-x^2 \text{ is maximum when } x^2\text{ is minimum, i.e. at } x=0,
   $$
   we get
   $$
   f(0)=\sqrt{25-0}=5.
   $$
   
Thus, the answers are:
- Minimum (smaller $$x$$-value): $$( -5,0 )$$
- Minimum (larger $$x$$-value): $$( 5,0 )$$
- Maximum: $$( 0,5 )$$

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(b) For $$[ -5,0 )$$:

1. The interval is half-closed; $$x=-5$$ is included and $$x=0$$ is not.
2. At $$x=-5$$,
   $$
   f(-5)=0.
   $$
3. Although $$f(x)$$ increases on $$[ -5,0 )$$ and
   $$
   \lim_{x\to 0^-}f(x)=\sqrt{25-0}=5,
   $$
   the value $$5$$ is never attained because $$x=0$$ is not in the interval.

Thus, the answers are:
- Minimum: $$( -5,0 )$$
- Maximum: DNE

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(c) For $$( -5,5 )$$:

1. Since the endpoints $$x=-5$$ and $$x=5$$ are not included, even though
   $$
   f(-5)=f(5)=0,
   $$
   these values are not attained.
2. The maximum occurs at $$x=0$$ where
   $$
   f(0)=5.
   $$
3. Regarding the minimum, for any $$ \epsilon>0 $$ one can find $$x$$ close to $$-5$$ or $$5$$ such that $$f(x)$$ is arbitrarily close to $$0$$, but no $$x$$ in $$( -5,5 )$$ yields $$f(x)=0$$.

Thus, the answers are:
- Minimum: DNE
- Maximum: $$( 0,5 )$$

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(d) For $$[ 1,5 )$$:

1. The interval includes $$x=1$$ but not $$x=5$$.
2. At $$x=1$$,
   $$
   f(1)=\sqrt{25-1^2}=\sqrt{24}=2\sqrt{6}.
   $$
3. For $$x$$ in $$[1,5 )$$, notice that $$ f'(x)=-\frac{x}{\sqrt{25-x^2}} $$ is negative (for $$x>0$$), so the function is decreasing. Hence, the maximum value is attained at $$x=1$$ and decreases as $$x$$ increases.
4. Although the function values approach
   $$
   \lim_{x\to 5^-}\sqrt{25-x^2}=\sqrt{25-25}=0,
   $$
   the value $$0$$ is not attained by any $$x$$ in $$[1,5 )$$.

Thus, the answers are:
- Minimum: DNE
- Maximum: $$( 1,2\sqrt{6} )$$
(a) - Minimum (smaller $$x$$-value): $$(-5, 0)$$ - Minimum (larger $$x$$-value): $$(5, 0)$$ - Maximum: $$(0, 5)$$ (b) - Minimum: $$(-5, 0)$$ - Maximum: DNE (c) - Minimum: DNE - Maximum: $$(0, 5)$$ (d) - Minimum: DNE - Maximum: $$(1, 2\sqrt{6})$$

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