Question

answer the following questions for the function \\( f(x) = xsqrt{x^2 + 36} \\) defined on the interval \\( -4 \leq x \leq 7 \\). \\( f(x) \\) is concave down on the interval \\( x = \\) \\( \square \\) to \\( x = \\) \\( \square \\) \\( f(x) \\) is concave up on the interval \\( x = \\) \\( \square \\) to \\( x = \\) \\( \square \\) the inflection point for this function is at \\( x = \\) \\( \square \\) the minimum for this function occurs at \\( x = \\) \\( \square \\) the maximum for this function occurs at \\( x = \\) \\( \square \\) question help: video message instructor

Explanation:

Step1: Find the first derivative

We use the product rule \((uv)^\prime = u^\prime v + uv^\prime\), where \(u = x\) and \(v=\sqrt{x^{2}+36}=(x^{2} + 36)^{\frac{1}{2}}\).
The derivative of \(u=x\) is \(u^\prime=1\).
The derivative of \(v=(x^{2}+36)^{\frac{1}{2}}\) using the chain rule \((g(h(x)))^\prime=g^\prime(h(x))\cdot h^\prime(x)\) where \(g(t)=t^{\frac{1}{2}}\) and \(h(x)=x^{2}+36\). So \(g^\prime(t)=\frac{1}{2}t^{-\frac{1}{2}}\) and \(h^\prime(x) = 2x\). Then \(v^\prime=\frac{1}{2}(x^{2}+36)^{-\frac{1}{2}}\cdot2x=\frac{x}{\sqrt{x^{2}+36}}\).
By the product rule, \(f^\prime(x)=1\cdot\sqrt{x^{2}+36}+x\cdot\frac{x}{\sqrt{x^{2}+36}}=\frac{x^{2}+36+x^{2}}{\sqrt{x^{2}+36}}=\frac{2x^{2}+36}{\sqrt{x^{2}+36}}\).

Step2: Find the second derivative

We use the quotient rule \((\frac{u}{v})^\prime=\frac{u^\prime v - uv^\prime}{v^{2}}\), where \(u = 2x^{2}+36\) and \(v=\sqrt{x^{2}+36}=(x^{2}+36)^{\frac{1}{2}}\).
\(u^\prime = 4x\), \(v^\prime=\frac{x}{\sqrt{x^{2}+36}}\) (from Step 1).

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Step3: Analyze the second derivative for concavity

The denominator \((x^{2}+36)^{\frac{3}{2}}\) is always positive for all real \(x\) since \(x^{2}+36>0\) and raising a positive number to a positive power keeps it positive. The numerator is \(2x(x^{2}+54)\), and \(x^{2}+54>0\) for all real \(x\). So the sign of \(f^{\prime\prime}(x)\) is determined by the sign of \(x\).

• When \(x<0\), \(f^{\prime\prime}(x)<0\), so the function is concave down.
• When \(x > 0\), \(f^{\prime\prime}(x)>0\), so the function is concave up.
• When \(x = 0\), \(f^{\prime\prime}(x)=0\), this is the inflection point.

Step4: Find the minimum and maximum on the interval \(- 4\leq x\leq7\)

The function \(f(x)=x\sqrt{x^{2}+36}\). We can also analyze the first derivative \(f^\prime(x)=\frac{2x^{2}+36}{\sqrt{x^{2}+36}}\). The numerator \(2x^{2}+36>0\) for all real \(x\) and the denominator \(\sqrt{x^{2}+36}>0\) for all real \(x\), so \(f^\prime(x)>0\) for all real \(x\). This means the function is increasing on its entire domain.
On the interval \([-4,7]\):

• The minimum occurs at the left - most point \(x=-4\) because the function is increasing.
• The maximum occurs at the right - most point \(x = 7\) because the function is increasing.

Answer:

s:

• \(f(x)\) is concave down on the interval \(x=-4\) to \(x = 0\)
• \(f(x)\) is concave up on the interval \(x = 0\) to \(x=7\)
• The inflection point for this function is at \(x = 0\)
• The minimum for this function occurs at \(x=-4\)
• The maximum for this function occurs at \(x = 7\)