​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
asked 2021-01-28
​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
\(\displaystyle{\left({a}\right)}{\frac{{{1}}}{{{7}}}}\)
\(\displaystyle{\left({b}\right)}{\frac{{{1}}}{{{9}}}}\)
\(\displaystyle{\left({c}\right)}{\frac{{{1}}}{{{11}}}}\)
\(\displaystyle{\left({d}\right)}{\frac{{{1}}}{{{13}}}}\)

Answers (1)

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: \(\displaystyle{f}^{{−{1}}}′{\left({9}\right)}={\frac{{{1}}}{{{\left({f}'{\left({{f}^{{-{1}}}{\left({9}\right)}}\right)}\right)}}}}=\frac{{1}}{{{f}'{\left({2}\right)}}}={\frac{{{1}}}{{{11}}}}\)
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