10. determine where the function is concave upward and where it is concave downward. use interval notation…

10.

Explanation:

Step1: Find the first - derivative

Using the power rule $(x^n)'=nx^{n - 1}$, for $f(x)=3x^{4}-30x^{3}+x - 5$, we have $f'(x)=12x^{3}-90x^{2}+1$.

Step2: Find the second - derivative

Differentiate $f'(x)$ again. $f''(x)=36x^{2}-180x=36x(x - 5)$.

Step3: Find the inflection points

Set $f''(x)=0$. So $36x(x - 5)=0$, which gives $x = 0$ and $x = 5$.

Step4: Test the intervals

Choose test points in the intervals $(-\infty,0)$, $(0,5)$ and $(5,\infty)$. For $x=-1$, $f''(-1)=36\times(-1)\times(-1 - 5)=216>0$, so $f(x)$ is concave upward on $(-\infty,0)$. For $x = 1$, $f''(1)=36\times1\times(1 - 5)=-144<0$, so $f(x)$ is concave downward on $(0,5)$. For $x = 6$, $f''(6)=36\times6\times(6 - 5)=216>0$, so $f(x)$ is concave upward on $(5,\infty)$.

Answer:

Concave upward: $(-\infty,0)\cup(5,\infty)$; Concave downward: $(0,5)$

11.

Explanation:

Step1: Find the first - derivative

For $g(x)=4x^{4}-8x^{3}+7$, using the power rule, $g'(x)=16x^{3}-24x^{2}$.

Step2: Find the second - derivative

Differentiate $g'(x)$: $g''(x)=48x^{2}-48x=48x(x - 1)$.

Step3: Find the inflection points

Set $g''(x)=0$. Then $48x(x - 1)=0$, which gives $x = 0$ and $x = 1$. Substitute $x = 0$ and $x = 1$ into $g(x)$: $g(0)=4\times0^{4}-8\times0^{3}+7 = 7$ and $g(1)=4\times1^{4}-8\times1^{3}+7=3$.

Answer:

The inflection points are $(0,7)$ and $(1,3)$

12.

Explanation:

Step1: Find the derivative

For $g(x)=\frac{1}{16}x^{2}-16\sqrt{x}=\frac{1}{16}x^{2}-16x^{\frac{1}{2}}$, using the power rule, $g'(x)=\frac{1}{8}x-\frac{8}{\sqrt{x}}=\frac{x^{\frac{3}{2}}-64}{8\sqrt{x}}$.

Step2: Find the critical points

Set $g'(x)=0$, then $x^{\frac{3}{2}}-64 = 0$, so $x^{\frac{3}{2}}=64$ and $x = 16$. Also, $g'(x)$ is undefined at $x = 0$.

Step3: Evaluate the function at critical and endpoints

$g(0)=\frac{1}{16}\times0^{2}-16\sqrt{0}=0$. $g(16)=\frac{1}{16}\times16^{2}-16\sqrt{16}=16 - 64=-48$. $g(25)=\frac{1}{16}\times25^{2}-16\sqrt{25}=\frac{625}{16}-80=\frac{625 - 1280}{16}=-\frac{655}{16}\approx - 40.94$.

Answer:

Absolute maximum value is $0$ at $x = 0$; Absolute minimum value is $-48$ at $x = 16$

13.

Explanation:

Step1: Let the dimensions of the rectangle

Let the length be $x$ and the width be $y$. The perimeter $P = 2x+2y=120$, so $y = 60 - x$.

Step2: Express the area function

The area $A=xy=x(60 - x)=60x - x^{2}$.

Step3: Find the derivative of the area function

$A'(x)=60 - 2x$.

Step4: Find the critical point

Set $A'(x)=0$, then $60 - 2x=0$, which gives $x = 30$.

Step5: Find the second - derivative

$A''(x)=-2<0$, so $x = 30$ is a maximum. When $x = 30$, $y=60 - 30 = 30$.

Answer:

The dimensions are $30$ ft by $30$ ft.