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Let $f:R\rightarrow R$ and $g:R\rightarrow R$ and $h:R\rightarrow R$ be differentiable function such that $f(x)=x^3+3x+2$ and $g(f(x))=x$ and $ h(g(g(x)))=x$ for all $x\in R$. Then which one is/are True

$\displaystyle (a) \ g'(2)=\frac{1}{15}$

$\displaystyle (b) \ h'(1)=666$

$\displaystyle (c) \ h(0)=16$

$\displaystyle (d) \ h(g(3))=36$

From $\displaystyle g(f(x))=x$

$\displaystyle g'(f(x))\cdot f'(x)=1$

Put $x=0,$ we have $\displaystyle g'(f(0))\cdot f'(0)=1$ And from

$f(x)=x^3+3x+2\Longrightarrow f(0)=2$ and $\displaystyle f'(x)=3x^2+3\Longrightarrow f'(0)=3$

So $\displaystyle g'(2)\cdot (3)=1\Longrightarrow g'(2)=\frac{1}{3}$

I did not understand how I find for $(b),(c),(d)$ parts

Please have a look

Priti Bisht
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  • JEE preparation question, by chance? I think it's important to mention your source, particularly because people search for questions by source. Edit your source into the question. If it's a textbook, add the textbook name , section number, etc. Similarly if it's from a PDF. If it is from a class then there is no need to add a source, but others will search for your question and stating the source makes it much easier. Thank you. – Sarvesh Ravichandran Iyer Mar 08 '23 at 06:17

2 Answers2

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For b), notice that if $h(g(g(x)))=x$ and $g(f(x))=x$, then you can do $x\to f(x)$ in the first of these expressions to get $h(g(g(f(x))))=f(x)$. This becomes $h(g(x))=f(x)$. Doing this again, we get $h(x)=f(f(x))$. Using $f(x)=x^3+3x+2$, this becomes: $$h(x)=(x^3+3x+2)^3+3(x^3+3x+2)+2.$$ And differentiating, we get $$h'(x)=3(x^3+3x+2)^2(3x^2+3)+3(3x^2+3)\implies h'(1)=3(1+3+2)^2(3+3)+3(3+3)=666.$$ So this one is true.

I think that with this, you should be able to solve c) and d).

Math Enjoyer
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    Thanks Math Enjoyer , $h(g(g(x))=x$, replace $x\rightarrow f(x),$ we have $h(g(g(f(x)))=f(x)\Longrightarrow h(g(x))=f(x)$ – Priti Bisht Mar 08 '23 at 07:30
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It is very obvious to show. Observe that $g$ is inverse of $f$. Then, for all $x$, $g(x) =f^{-1}(x) $ Now, we have to find $h(0)$

Let, $g(g(a))=0$

$g(a) =g^{-1}(0)$

$g(a) =f(0)$

$g(a) =2$

$a=g^{-1}(2) $

$a=f(2)=8+6+2=16$

$h(0) =16$

In this way, I hope you can show other two, just use the fact $g$ is inverse of $f$.

Nope
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