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Find higher derivatives of rational and radical functions

\begin{cases} f(1)=-3 \ f(n)=2\cdot f(n-1)+1 \end{cases}

f(2)=

Subtract mixed fraction:

\frac{12(2)}{8}-\frac{10(5)}{6}

Add the functions: \begin{align*} &f(x)=3x-5x^{2}-x^{3},\ &g(x)=6x^{2}-4x \end{align*}

The derivative of a function

f

f^{\prime}(x)=e^{x}-x^{3}

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On which intervals is the graph of

f

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Use a graphing calculator.

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1

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[1.857,4.536]

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(-\infty, 1.857]

[4.536, \infty)

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(-\infty,-0.459]

[0.91,3.733]

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[-0.459,0.91]

[3.733, \infty)

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(E) All real numbers

The second derivative of a function

f

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f^{\prime \prime}(x)=\sin \left(x^{2}\right)+\cos (2 x)-\frac{1}{2} \text {. }

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How many points of inflection does the graph of

f

have on the interval

1<x<4

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Use a graphing calculator.

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1

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The second derivative of a function

h

h^{\prime \prime}(x)=\frac{\sqrt[3]{x}}{e^{x}}+1

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Where does the graph of

h

have a point of inflection?

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Use a graphing calculator.

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1

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x=-0.35

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x=0

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x=0.33

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x=1.497

\begin{array}{l}f^{\prime}(x)=\frac{81}{x^{4}} \text { and } f(-1)=18 . \\ f(3)=\end{array}

\begin{array}{l}f^{\prime}(x)=\frac{16}{x^{2}} \text { and } f(-2)=0 . \\ f(4)=\square\end{array}

\begin{array}{l}f^{\prime}(x)=-\frac{6}{x^{4}} \text { and } f(1)=-4 . \\ f(-1)=\end{array}

\begin{array}{l}f^{\prime}(x)=\frac{24}{x^{3}} \text { and } f(2)=12 . \\ f(-1)=\end{array}

\begin{array}{l}G(x)=(x+3)^{2} \\ g(x)=G^{\prime}(x) \\ \int_{-1}^{7} g(x) d x=\end{array}

F(x)=4 x^{2}-5 x

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\begin{array}{l}f(x)=F^{\prime}(x) \\ \int_{1}^{4} f(x) d x=\end{array}

n

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\frac{12}{5}=\frac{n}{8}

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n =\square

\begin{array}{l}x^{2}+x-72=0 \\ (x+\square)^{2}=\square\end{array}

Find the missing term.

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(x+2)(x+5)

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x^{2}+5 x+2 x+(2) \cdot(5)

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x^{2}\_\_\_\_\_+10

z_{1} z_{2}

. Write the answer in standard form.

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\begin{array}{l} z_{1}=7 \operatorname{cis} \frac{\pi}{3} \\ z_{2}=2 \operatorname{cis} \frac{11 \pi}{6} \end{array}

\left\{\begin{array}{l} b(1)=-500 \\ b(n)=b(n-1) \cdot \frac{4}{5} \end{array}\right.

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3^{\text {rd }}

term in the sequence?

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\square

\begin{array}{l}10^{x+2 y}=5 \\ 10^{3 x+4 y}=50\end{array}

\frac{2}{x+1}+\frac{x}{x-1}=\frac{2}{x^{2}-1}

T(x) = \frac{e^{x^2 + 8x}}{\sqrt{x^4 + 7}},\quad T'(x) =

Complete the recursive formula of the arithmetic sequence

-15, -11, -7, -3, \dots

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c(1)=\square

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c(n)=c(n-1)+\square

\begin{array}{r}4 x_{2}+x_{3}=0 \\ x_{1}+x_{2}+x_{3}=0 \\ 2 x_{1}+4 x_{2}+3 x_{3}=0\end{array}

\begin{array}{l}h(x)=x^{2}+x \\ g(x)=3 x+5 \\ \text{ Find }(h \cdot g)(x)\end{array}

\begin{array}{c}x^{2}-6 x+11=y \\ x=y+1\end{array}

\begin{array}{l}f^{\prime}(x)=12 e^{x} \text { and } f(4)=-16+12 e^{4} . \\ f(0)=\end{array}

n

\frac{12}{5}=\frac{n}{8}

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n = \Box

Solve the equation. \begin{align*} \frac{b}{6} &= 3, \ b &= \Box \end{align*}

Solve the equation.

\begin{align*} \frac{b}{6} &= 3 \\ b &= \Box \end{align*}

n

\begin{array}{l}\frac{8}{3}=\frac{12}{n} \\ n=\square\end{array}

\begin{array}{l}f^{\prime}(x)=-5 e^{x} \text { and } f(3)=22-5 e^{3} . \\ f(0)=\square\end{array}

\begin{array}{l}f^{\prime}(x)=-27 e^{x} \text { and } f(6)=36-27 e^{6} . \\ f(0)=\square\end{array}

\begin{array}{l}f^{\prime}(x)=12 x^{2}-6 x+2 \text { and } f(-1)=3 . \\ f(2)=\end{array}

\begin{array}{l}f^{\prime}(x)=5 e^{x} \text { and } f(7)=40+5 e^{7} . \\ f(0)=\square\end{array}

\begin{array}{l}f^{\prime}(x)=12 e^{x} \text { and } f(4)=-16+12 e^{4} . \\ f(0)=\end{array}

\begin{array}{l}f(x)=x^{11} \\ f^{\prime}(x)=\end{array}

\frac{d^{2}}{d x^{2}}[2 \sin (-4 x-3)]

\begin{aligned} h(x) & =(5-6 x)^{5} \\ h^{\prime}(x) & =? \end{aligned}

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1

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-6 x^{5}+5 x^{4}(5-6 x)

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-30(5-6 x)^{4}

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(-6)^{5}

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5(5-6 x)^{4}