Answer the following questions for the function

defined on the interval .
is concave down on the interval
to
is concave up on the interval
to
The inflection point for this function is at
The minimum for this function occurs at
The maximum for this function occurs at
T

Ask by Ortiz Jimenez. in the United States

Mar 25,2025

Question

Answer the following questions for the function

defined on the interval .
is concave down on the interval
to
is concave up on the interval
to
The inflection point for this function is at
The minimum for this function occurs at
The maximum for this function occurs at
T

Ask by Ortiz Jimenez. in the United States

Mar 25,2025

UpStudy AI · Accepted Answer

**Step 1. Find the first derivative $$ f'(x) $$.** We have $$ f(x)=x\sqrt{x^2+25}. $$ Using the product rule: $$ f'(x)=\frac{d}{dx}\Bigl[x\Bigr]\sqrt{x^2+25}+x\frac{d}{dx}\Bigl[\sqrt{x^2+25}\Bigr]. $$ Since $$ \frac{d}{dx}[x]=1, $$ and using the chain rule for $$ \sqrt{x^2+25}=(x^2+25)^{1/2}, $$ we have $$ \frac{d}{dx}\sqrt{x^2+25}=\frac{1}{2}(x^2+25)^{-1/2}(2x)=\frac{x}{\sqrt{x^2+25}}. $$ Thus, $$ f'(x)=\sqrt{x^2+25}+x\cdot\frac{x}{\sqrt{x^2+25}}=\sqrt{x^2+25}+\frac{x^2}{\sqrt{x^2+25}}. $$ To combine these, write: $$ f'(x)=\frac{x^2+25+x^2}{\sqrt{x^2+25}}=\frac{2x^2+25}{\sqrt{x^2+25}}. $$ --- **Step 2. Find the second derivative $$ f''(x) $$.** We have $$ f'(x)=\frac{2x^2+25}{\sqrt{x^2+25}}. $$ Let $$ N(x)=2x^2+25 $$ and $$ D(x)=\sqrt{x^2+25}=(x^2+25)^{1/2} $$. Then, $$ f''(x)=\frac{N'(x)D(x)-N(x)D'(x)}{[D(x)]^2}. $$ Differentiate the numerator and denominator: $$ N'(x)=4x, $$ and as before, $$ D'(x)=\frac{x}{\sqrt{x^2+25}}. $$ Since $$ [D(x)]^2=x^2+25, $$ we have $$ f''(x)=\frac{4x\sqrt{x^2+25}-\left(2x^2+25\right)\frac{x}{\sqrt{x^2+25}}}{x^2+25}. $$ To combine the terms in the numerator, obtain a common denominator: $$ 4x\sqrt{x^2+25}=\frac{4x(x^2+25)}{\sqrt{x^2+25}}, $$ so $$ f''(x)=\frac{\frac{4x(x^2+25)-x(2x^2+25)}{\sqrt{x^2+25}}}{x^2+25}=\frac{4x(x^2+25)-x(2x^2+25)}{(x^2+25)^{3/2}}. $$ Simplify the numerator: $$ 4x(x^2+25)=4x^3+100x, $$ $$ x(2x^2+25)=2x^3+25x, $$ so $$ 4x^3+100x-2x^3-25x=2x^3+75x=x(2x^2+75). $$ Thus, $$ f''(x)=\frac{x(2x^2+75)}{(x^2+25)^{3/2}}. $$ --- **Step 3. Determine the intervals of concavity.** The denominator $$(x^2+25)^{3/2}$$ is always positive. Therefore, the sign of $$ f''(x) $$ is determined by the numerator $$ x(2x^2+75) $$. Since $$ 2x^2+75>0 \quad \text{for all } x, $$ the sign of $$ f''(x) $$ is the same as the sign of $$ x $$. - For $$ x<0 $$: $$ f''(x)<0 $$ (concave down). - For $$ x>0 $$: $$ f''(x)>0 $$ (concave up). Thus: - $$ f(x) $$ is concave down on the interval from $$ x=-6 $$ to $$ x=0 $$. - $$ f(x) $$ is concave up on the interval from $$ x=0 $$ to $$ x=6 $$. The inflection point occurs where $$ f''(x)=0 $$, which is when $$ x=0 $$. --- **Step 4. Find the minimum and maximum on $$ -6\le x\le6 $$.** Since $$ f'(x)=\frac{2x^2+25}{\sqrt{x^2+25}}>0 \quad \text{for all } x, $$ $$ f(x) $$ is increasing on the entire interval. Therefore: - The minimum occurs at the left endpoint $$ x=-6 $$. - The maximum occurs at the right endpoint $$ x=6 $$. --- **Final Answers:** - $$ f(x) $$ is concave down on the interval $$ x=-6 $$ to $$ x=0 $$. - $$ f(x) $$ is concave up on the interval $$ x=0 $$ to $$ x=6 $$. - The inflection point for this function is at $$ x=0 $$. - The minimum for this function occurs at $$ x=-6 $$. - The maximum for this function occurs at $$ x=6 $$. - $$ f(x) $$ is concave down from $$ x=-6 $$ to $$ x=0 $$. - $$ f(x) $$ is concave up from $$ x=0 $$ to $$ x=6 $$. - The inflection point is at $$ x=0 $$. - The minimum occurs at $$ x=-6 $$. - The maximum occurs at $$ x=6 $$.

Answer the following questions for the function

defined on the interval .
is concave down on the interval
to
is concave up on the interval
to
The inflection point for this function is at
The minimum for this function occurs at
The maximum for this function occurs at
T

Solución de inteligencia artificial de Upstudy

Responder

Solución

Bonus Knowledge

Concavity:

Inflection Point:

Extrema:

preguntas relacionadas

Latest Calculus Questions

Answer the following questions for the function defined on the interval . is concave down on the interval to is concave up on the interval to The inflection point for this function is at The minimum for this function occurs at The maximum for this function occurs at T

Solución de inteligencia artificial de Upstudy

Responder

Solución

Bonus Knowledge

Concavity:

Inflection Point:

Extrema:

preguntas relacionadas

Latest Calculus Questions

Answer the following questions for the function

defined on the interval .
is concave down on the interval
to
is concave up on the interval
to
The inflection point for this function is at
The minimum for this function occurs at
The maximum for this function occurs at
T