# ​If f(x)= frac{3}{4x^{3}}+2x−1 then the value of (f−1)′(x)(f^{−1})′(x) when x=9 (a)frac{1}{7} (b)frac{1}{9} (c)frac{1}{11} (d)frac{1}{13}

Question
Functions
​If $$\displaystyle{f{{\left({x}\right)}}}={\frac{{{3}}}{{{4}{x}^{{{3}}}}}}+{2}{x}−{1}$$ then the value of $$\displaystyle{\left({f}−{1}\right)}′{\left({x}\right)}{\left({f}^{{−{1}}}\right)}′{\left({x}\right)}$$ when x=9
$$\displaystyle{\left({a}\right)}{\frac{{{1}}}{{{7}}}}$$
$$\displaystyle{\left({b}\right)}{\frac{{{1}}}{{{9}}}}$$
$$\displaystyle{\left({c}\right)}{\frac{{{1}}}{{{11}}}}$$
$$\displaystyle{\left({d}\right)}{\frac{{{1}}}{{{13}}}}$$

2021-01-29
Here we are using the following derivative rule: $$\displaystyle{f}^{{−{1}}}′{\left({a}\right)}={\frac{{{1}}}{{{\left({f}'{\left({{f}^{{-{{1}}}}{\left({a}\right)}}\right)}\right\rbrace}{\left({f}'{\left({{f}^{{-{1}}}{\left({a}\right)}}\right)}\right.}}}}$$ where a=9. A couple of things to first look out for is the following: When f(b)=3/4b^3+2b−1=9, we have that f^−1(9)=b. So we must solve for the value ofb. Doing so, you will see that b=2b=2. Then, f′(2)=9/4*2^2+2=11. So our equation will then look as follows: $$\displaystyle{f}^{{−{1}}}′{\left({9}\right)}={\frac{{{1}}}{{{\left({f}'{\left({{f}^{{-{1}}}{\left({9}\right)}}\right)}\right)}}}}=\frac{{1}}{{{f}'{\left({2}\right)}}}={\frac{{{1}}}{{{11}}}}$$

### Relevant Questions

​If $$f(x)= \frac{3}{4x^{3}}+2x−1$$ then the value of $$(f−1)′(x)(f^{−1})′(x)$$ when x=9
$$(a)\frac{1}{7}$$
$$(b)\frac{1}{9}$$
$$(c)\frac{1}{11}$$
$$(d)\frac{1}{13}$$
Whch of the two functions below has the smallest minimum y-value?
$$f(x)=4(x-6)^4+1, g(x)=2x^3+28$$.
A.There is not enough information to determine.
B. g(x)
C.f(x)
D. The extreme minimum y-value for f(x)and g(x)is - infinity.
Determine the algebraic modeling which of the following data sets are linear and which are exponential. For the linear sets, determine the slope. For the exponential sets, determine the growth factor or the decay factor
a) $$\begin{array}{|c|c|}\hline x & -2 & -1 & 0 & 1 & 2 & 3 & 4 \\ \hline y & \frac{1}{9} & \frac{1}{3} & 1 & 3 & 9 & 27 & 81 \\ \hline \end{array}$$ b) $$\begin{array}{|c|c|}\hline x & -2 & -1 & 0 & 1 & 2 & 3 & 4 \\ \hline y & 2 & 2.6 & 3.2 & 3.8 & 4.4 & 5.0 & 5.6 \\ \hline \end{array}$$
c) $$\begin{array}{|c|c|}\hline x & -2 & -1 & 0 & 1 & 2 & 3 & 4 \\ \hline y & 3.00 & 5.0 & 7 & 9 & 11 & 13 & 15 \\ \hline \end{array}$$
d) $$\begin{array}{|c|c|}\hline x & -2 & -1 & 0 & 1 & 2 & 3 & 4 \\ \hline y & 5.25 & 2.1 & 0.84 & 0.336 & 0.1344 & 0.5376 & 0.021504 \\ \hline \end{array}$$
For what value of the constant c is the function f continuous on $$\displaystyle{\left(−∞,+∞\right)}?$$
$$\displaystyle{b}{e}{g}\in{\left\lbrace{a}{l}{i}{g}{n}\cdot\right\rbrace}$$ PSKf(x)&=\left{\begin{matrix}cx^2+4x &\text{ if } x\lt 5\ x^3-cx &\text{ if }x\geq 5\end{matrix}\right. \end{align*}ZSK c=
Suppose that f(x) is a continuous, one-to-one function such that $$\displaystyle{f{{\left({2}\right)}}}={1},{f}'{\left({2}\right)}=\frac{{1}}{{4}},{f{{\left({1}\right)}}}={3}$$, and f '(1) = 7. Let $$\displaystyle{g{{\left({x}\right)}}}={{f}^{{−{1}}}{\left({x}\right)}}$$ and let $$\displaystyle{G}{\left({x}\right)}={x}^{{2}}\cdot{g{{\left({x}\right)}}}$$. Find G'(1). (You may not need to use all of the provided information.)
Gastroenterology
We present data relating protein concentration to pancreatic function as measured by trypsin secretion among patients with cystic fibrosis.
If we do not want to assume normality for these distributions, then what statistical procedure can be used to compare the three groups?
Perform the test mentioned in Problem 12.42 and report a p-value. How do your results compare with a parametric analysis of the data?
Relationship between protein concentration $$(mg/mL)$$ of duodenal secretions to pancreatic function as measured by trypsin secretion:
$$\left[U/\left(k\ \frac{g}{h}r\right)\right]$$
Tapsin secreton [UGA]
$$\leq\ 50$$
$$\begin{array}{|c|c|}\hline \text{Subject number} & \text{Protetion concentration} \\ \hline 1 & 1.7 \\ \hline 2 & 2.0 \\ \hline 3 & 2.0 \\ \hline 4 & 2.2 \\ \hline 5 & 4.0 \\ \hline 6 & 4.0 \\ \hline 7 & 5.0 \\ \hline 8 & 6.7 \\ \hline 9 & 7.8 \\ \hline \end{array}$$
$$51\ -\ 1000$$
$$\begin{array}{|c|c|}\hline \text{Subject number} & \text{Protetion concentration} \\ \hline 1 & 1.4 \\ \hline 2 & 2.4 \\ \hline 3 & 2.4 \\ \hline 4 & 3.3 \\ \hline 5 & 4.4 \\ \hline 6 & 4.7 \\ \hline 7 & 6.7 \\ \hline 8 & 7.9 \\ \hline 9 & 9.5 \\ \hline 10 & 11.7 \\ \hline \end{array}$$
$$>\ 1000$$
$$\begin{array}{|c|c|}\hline \text{Subject number} & \text{Protetion concentration} \\ \hline 1 & 2.9 \\ \hline 2 & 3.8 \\ \hline 3 & 4.4 \\ \hline 4 & 4.7 \\ \hline 5 & 5.5 \\ \hline 6 & 5.6 \\ \hline 7 & 7.4 \\ \hline 8 & 9.4 \\ \hline 9 & 10.3 \\ \hline \end{array}$$
Solve the equations and inequalities below, if possible. $$\displaystyle{a}.\sqrt{{{x}−{1}}}+{13}={13}$$
$$\displaystyle{b}.{6}{\left|{x}\right|}{>}{18}$$
$$\displaystyle{c}.{\left|{3}{x}-{2}\right|}\le{2}$$
$$\displaystyle{d}.{\frac{{{4}}}{{{5}}}}-{\frac{{{2}{x}}}{{{3}}}}={\frac{{{3}}}{{{10}}}}$$
$$\displaystyle{e}.{\left({4}{x}-{2}\right)}^{{{2}}}\le{100}$$
$$\displaystyle{f}.{\left({x}-{1}\right)}^{{{3}}}={8}$$
Lesson 3−23 - 23−2
Some Attributes of Polynomial Functions
$$\displaystyle{a}.{f{{\left({x}\right)}}}={5}{x}−{x}{3}+{3}{x}{5}−{2}{f{{\left({x}\right)}}}={5}{x}-{x}^{{{3}}}+{3}{x}^{{{5}}}-{2}{f{{\left({x}\right)}}}={5}{x}−{x}{3}+{3}{x}{5}−{2}$$
$$\displaystyle{b}.{f{{\left({x}\right)}}}=−{22}{x}{3}−{8}{x}{4}−{2}{x}+{7}{f{{\left({x}\right)}}}=-{\frac{{{2}}}{{{2}}}}{x}^{{{3}}}-{8}{x}^{{{4}}}-{2}{x}+{7}{f{{\left({x}\right)}}}=−{22}​{x}{3}−{8}{x}{4}−{2}{x}+{7}$$
$$A. (0,8), (3,8), (1,6)$$
$$B. (4,2), (6,1), (8,9)$$
$$C. (1,20), (2,23), (9,26)$$
$$D. (0,3), (2,3), (2,0)$$
$$\displaystyle{A}.{\left({0},{8}\right)},{\left({3},{8}\right)},{\left({1},{6}\right)}$$
$$\displaystyle{B}.{\left({4},{2}\right)},{\left({6},{1}\right)},{\left({8},{9}\right)}$$
$$\displaystyle{C}.{\left({1},{20}\right)},{\left({2},{23}\right)},{\left({9},{26}\right)}$$
$$\displaystyle{D}.{\left({0},{3}\right)},{\left({2},{3}\right)},{\left({2},{0}\right)}$$