Animated Solution for Mathematics - Differentiation: Let $f:R\to R,g:R\to R$ and $h:R\to R$ be differentiable functions such that $f(x)=x^3+3x+2,g(f(x))=x$ and $h(g(g(x)))=x$ for all $x\in R$. Then

Imagine you are standing before a vast, shimmering mirror. In mathematics, the line $y=x$ is that mirror. When we talk about inverse functions, we are talking about a reflection across this very line.

In this problem, we are given $f(x)=x^3+3x+2$ and the relation $g(f(x))=x$. This is the classic, textbook definition of an inverse function, where $g=f^{-1}$.

Geometrically, if you were to plot the curve of $f(x)$ in blue, its reflection across the line $y=x$ would give you the curve of $g(x)$ in green. They are perfectly symmetric, a beautiful dance of coordinates where every point $(x,y)$ on $f$ becomes $(y,x)$ on $g$.

A Shortcut to Mastery

Now, we are asked to find $g^{\prime }(2)$. A common trap is to try and find the explicit formula for $g(x)$, but solving the cubic $f(x)=x^3+3x+2$ for $x$ is a nightmare.

Instead, we use the inverse derivative rule:

where $y=f(x)$. This rule tells us that the slope of the inverse function at $y$ is simply the reciprocal of the slope of the original function at $x$.

First, we find the $x$ that maps to $y=2$. Setting $x^3+3x+2=2$, we get $x^3+3x=0$, which simplifies to $x(x^2+3)=0$. The only real solution is $x=0$.

Now, we find the derivative of $f(x)$:

At $x=0$, $f^{\prime }(0)=3$. Therefore, $g^{\prime }(2)=\frac{1}{f^{\prime }(0)}=\frac{1}{3}$. Since $\frac{1}{3} eq \frac{1}{15}$, we can confidently say that Option 1 is false.

Next, we encounter the function $h(x)$ defined by $h(g(g(x)))=x$. This looks intimidating, but we can break it down. We know that $f$ and $g$ are inverses, meaning they undo each other.

If we apply the function $f$ to both sides of the equation $h(g(g(x)))=x$, the outermost $g$ is cancelled out, leaving us with $h(g(x))=f(x)$. If we apply $f$ once more, the remaining $g$ is cancelled, and we arrive at the simplified relation:

Now, checking Option 3, we calculate $h(0)=f(f(0))$. First, $f(0)=0^3+3(0)+2=2$. Then, $f(2)=2^3+3(2)+2=16$. Thus, $h(0)=16$, and Option 3 is true.

The Engine of Calculus

Finally, we tackle Option 2: $h^{\prime }(1)$. We have $h(x)=f(f(x))$. To differentiate this, we use the chain rule:

This is the engine of calculus, allowing us to differentiate composite functions with ease. Substituting $x=1$, we get $h^{\prime }(1)=f^{\prime }(f(1))\cdot f^{\prime }(1)$.

First, $f(1)=1^3+3(1)+2=6$. Next, $f^{\prime }(1)=3(1{)}^2+3=6$. Finally, $f^{\prime }(f(1))=f^{\prime }(6)=3(6{)}^2+3=111$.

Multiplying these together:

Option 2 is true! We have navigated the geometry of inverses, mastered the inverse derivative rule, and wielded the chain rule to uncover the secrets of composite functions. You have successfully decoded the problem.