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Inverse function Quiz 2

mathematics

 Find the inverse of the function $f\left( x \right)={{4}^{3x+2}}$

 First take $y={{4}^{3x+2}}$ and express $x\,$  in terms of  $y$  and then replace  $y$   with $x\,$ to obtain the inverse of the given function.

Let, $y={{4}^{3x+2}}$ $ \\ $ $\Rightarrow 3x+2={{\log }_{4}}\left( y \right)$ $ \\ $ $\Rightarrow 3x={{\log }_{4}}\left( y \right)-2$ $ \\ $ $={{\log }_{4}}\left( y \right)-{{\log }_{4}}\left( 16 \right)$ $ \\ $ $={{\log }_{4}}\left( \dfrac{y}{16} \right)$ $ \\ $ $\Rightarrow x\,\,=\dfrac{1}{3}{{\log }_{4}}\left( \dfrac{y}{16} \right)\,\,$ $ \\ $ $\therefore {{f}^{-1}}\left( x \right)\,=\dfrac{1}{3}{{\log }_{4}}\left( \dfrac{x}{16} \right)\,\,$

 If ${{\sin }^{-1}}x-{{\cos }^{-1}}\sqrt{1-{{x}^{2}}}=\dfrac{-\pi }{2}$, then find x

 Substitute $x=\sin \theta $  in the given equation and then find the value of   $\theta $  .Finally find the value of  $x$.

 Let us assume, $x=\sin \theta $ $ \\ $ $\Rightarrow {{\sin }^{-1}}\left( \sin \theta  \right)-{{\cos }^{-1}}\left( \sqrt{1-{{\sin }^{2}}\theta } \right)=\dfrac{-\pi }{2}$ $ \\ $ $\Rightarrow \theta -{{\cos }^{-1}}\left( \cos \theta  \right)=\dfrac{-\pi }{2}$ $ \\ $ $\Rightarrow \theta -{{\cos }^{-1}}\left( \cos \left( -\theta  \right) \right)=\dfrac{-\pi }{2}$ $ \\ $ $\Rightarrow \theta -\left( -\theta  \right)=\dfrac{-\pi }{2}$ $ \\ $ $\theta =\dfrac{-\pi }{4}$ $ \\ $ Hence, $x=\sin \left( \dfrac{-\pi }{4} \right)=\dfrac{-1}{\sqrt{2}}$

 The function  $f$ is not a one-to-one function. Restrict its domain until it does become one-to-one . The inverse of this function is ${{f}^{-1}}$ . Consider the following statements:  $ \\ $ Statement 1: ${{f}^{-1}}\left( f\left( x \right) \right)=x$ $ \\ $ Statement 2: $f\left( {{f}^{-1}}\left( x \right) \right)=x$

 Choose a function as $f\left( x \right)$ and then restrict its domain to make it one-to-one thereby verifying the given statements.

 Let $f\left( x \right)={{x}^{2}}$. $ \\ $ Reduce its domain to $\left[ 0,\infty  \right]$ and then the inverse function is ${{f}^{-1}}\left( x \right)=\sqrt{x}$ $ \\ $ Consider Statement 1 at $x=-3$ $ \\ $  ${{f}^{-1}}\left( f\left( -3 \right) \right)=\sqrt{{{\left( -3 \right)}^{2}}}=\sqrt{9}=3$ $ \\ $ We started with $-3$ and landed on $3$. $ \\ $ Hence , ${{f}^{-1}}\left( f\left( x \right) \right)=\sqrt{{{x}^{^{2}}}}=x$ is false , i.e Statement 1 is false $ \\ $ But , $f\left( {{f}^{-1}}\left( x \right) \right)={{\left( \sqrt{x} \right)}^{2}}=x$ is still true , i.e Statement 2 is true 

 Let $f:\mathbb{R}\to \left[ -1,1 \right]$ and $f\left( x \right)=\cos x$; where $f$ is not an invertible function. Its domain is restricted to $\left[ -\pi ,0 \right]$ so that it is invertible, then ${{f}^{-1}}:\left[ -1,1 \right]\to \left[ -\pi ,0 \right]$is the inverse mapping. The value of $-\left( {{f}^{-1}}\left( \cos 10 \right) \right)$ has its integral part equal to 

 Using the inequality $3\pi <10<\left( 3\pi +\dfrac{\pi }{2} \right)<4\pi $ , obtain the relation between ${{f}^{-1}}\left( \cos 10 \right)$ and ${{f}^{-1}}\left( \cos \left( 10-4\pi  \right) \right)$ and then solve further to obtain the answer.

It is a one-one function. $ \\ $ We know that,  $ \\ $ $3\pi <10<\left( 3\pi +\dfrac{\pi }{2} \right)<4\pi $ $ \\ $ $\Rightarrow -\pi <10-4\pi <0$ $ \\ $ Now, $-{{f}^{-1}}\left( \cos 10 \right)=-{{f}^{-1}}\left( \cos \left( 10-4\pi  \right) \right)$ $ \\ $ $=-\left( 10-4\pi  \right)=-{{f}^{-1}}\left( \cos 10 \right)=4\pi -10$ $ \\ $ $\therefore \left[ 4\pi -10 \right]=2$

  If $f(x)=\cos x+3x$and ${{f}^{-1}}\left( \dfrac{3\pi }{2} \right)=\dfrac{\pi }{2}$, find the derivative of ${{f}^{-1}}$ at $x=\dfrac{3\pi }{2}$.

 By using inverse function theorem ,express ${{\left( {{f}^{-1}} \right)}^{\prime }}\left( \dfrac{3\pi }{2} \right)$ in terms of ${f}'\left( \dfrac{\pi }{2} \right)$ and then find the value of ${f}'\left( \dfrac{\pi }{2} \right)$ to obtain the answer. 

We have , $f(x)=\cos x+3x$ $ \\ $ ${f}'(x)=-\sin x+3$ $ \\ $ We know that $-1\le -\sin x\le 1$ $ \\ $ $2\le -\sin x+3\le 4$ $ \\ $ Hence, $f(x)$ is an increasing function. $ \\ $ By using Inverse function theorem, $ \\ $ ${{\left( {{f}^{-1}} \right)}^{\prime }}\left( x \right)=\dfrac{1}{{f}'\left( {{f}^{-1}}\left( x \right) \right)}$ $ \\ $ Put $x=\dfrac{3\pi }{2}$ $ \\ $ ${f}'\left( {{f}^{-1}}\left( x \right) \right)={f}'\left( {{f}^{-1}}\left( \dfrac{3\pi }{2} \right) \right)$   $\Rightarrow {f}'\left( {{f}^{-1}}\left( \dfrac{3\pi }{2} \right) \right)={f}'\left( \dfrac{\pi }{2} \right)$ $ \\ $ As we know that ${f}'(x)=-\sin x+3$ $ \\ $ ${f}'\left( \dfrac{\pi }{2} \right)=-\sin \left( \dfrac{\pi }{2} \right)+3$   $\Rightarrow {f}'\left( \dfrac{\pi }{2} \right)=-1+3=2$ $ \\ $ $\therefore {{\left( {{f}^{-1}} \right)}^{\prime }}\left( \dfrac{3\pi }{2} \right)=\dfrac{1}{2}$