The inverse of $x+x^3$ ¶

Math I: Calculus · Week 5 test, open question · 7 points · Lectures 7–8

Problem¶

Define $f\colon\mathbb{R}\to\mathbb{R}$ by $f(x)=x+x^3\text{.}$

Prove that $f$ is invertible from $\mathbb{R}$ onto $\mathbb{R}\text{.}$ [2 points]
Let $g=f^{-1}\text{.}$ Find $g'(0)$ and $g'(2)\text{.}$ [2 points]
Evaluate [3 points]

$\lim_{y\to0}\frac{g(y)-y}{y^3}.$

What you need¶

Bijection and inverse function: invertible means injective and surjective.
Monotone function: the sign test $f'>0\Rightarrow f$ strictly increasing.
Intermediate value theorem and limits at infinity, for surjectivity.
Inverse function theorem: $g'(y)=1/f'(g(y))\text{.}$
For part 3, any one of: substitution in a limit, the definition of the derivative, or L'Hôpital's rule.

Plan¶

There is no formula for $g\text{,}$ and you do not need one. Everything comes from the relation $f(g(y))=y\text{.}$ For values of $g$ at specific points, solve $f(x)=y$ by inspection. For the limit, either rewrite everything in terms of $x=g(y)\text{,}$ where it becomes a simple fraction, or use $g(y)/y\to g'(0)=1\text{.}$

Step-by-step solution¶

Try each step yourself before you open it.

Part 1, Step 1 · Injective

$f'(x)=1+3x^2>0$ for every $x\text{,}$ so $f$ is strictly increasing by the sign test, hence injective.

Concept: Monotone function

Part 1, Step 2 · Surjective (with Step 1: 2 points)

$f$ is continuous, $f(x)\to-\infty$ as $x\to-\infty$ and $f(x)\to+\infty$ as $x\to+\infty\text{.}$ For any $y\in\mathbb{R}$ pick $a<b$ with $f(a)<y<f(b)\text{;}$ the intermediate value theorem gives $x\in(a,b)$ with $f(x)=y\text{.}$ So the range is all of $\mathbb{R}$ and $f$ is invertible.

Concept: Intermediate value theorem

Part 2 · Two derivatives of the inverse (2 points)

Since $f'(x)=1+3x^2\neq0$ for every $x\text{,}$ the inverse function theorem applies:

g'(y)=\frac{1}{f'(g(y))}=\frac{1}{1+3g(y)^2}.

Find $g(0)$ and $g(2)$ by solving $f(x)=y\text{:}$ $f(0)=0$ gives $g(0)=0\text{,}$ and $f(1)=2$ gives $g(2)=1\text{.}$ Hence

g'(0)=\frac{1}{1}=1,\quad g'(2)=\frac{1}{1+3}=\frac{1}{4}.

Concept: Inverse function theorem

Part 3, Method 1 · Substitute x = g(y) (3 points)

Put $x=g(y)\text{.}$ Then $y=f(x)=x+x^3\text{.}$ Since $g$ is differentiable by Part 2, it is continuous, and $g(0)=0\text{;}$ so $x\to0$ as $y\to0\text{,}$ and $x\neq0$ when $y\neq0\text{.}$ So

\begin{aligned} \frac{g(y)-y}{y^3}&=\frac{x-(x+x^3)}{(x+x^3)^3}\\ &=-\frac{x^3}{x^3(1+x^2)^3}\\ &=-\frac{1}{(1+x^2)^3}\to-1. \end{aligned}

Part 3, Method 2 · Use g(y)/y → g′(0) (3 points)

From $f(g(y))=y$ we get $g(y)+g(y)^3=y\text{,}$ so $g(y)-y=-g(y)^3\text{.}$ For $y\neq0\text{,}$

\frac{g(y)-y}{y^3}=-\left(\frac{g(y)}{y}\right)^3.

Because $g(0)=0\text{,}$ the quotient $g(y)/y$ is a difference quotient:

\lim_{y\to0}\frac{g(y)-g(0)}{y-0}=g'(0)=1.

So the limit is $-(1)^3=-1\text{.}$ Substituting $y=0$ directly gives $0/0$ and earns nothing; the justification $g(y)/y\to1$ is the point.

Part 3, Method 3 · L'Hôpital's rule once (3 points)

The quotient is of type $0/0\text{.}$ Differentiate numerator and denominator, using Part 2:

\frac{g'(y)-1}{3y^2} =\frac{1}{3y^2}\left(\frac{1}{1+3g(y)^2}-1\right) =-\frac{(g(y)/y)^2}{1+3g(y)^2}\to-1,

since $g(y)/y\to1$ and $g(y)\to0\text{.}$ The derivative quotient has a limit, which justifies L'Hôpital's rule, so the original limit is also $-1\text{.}$

Concept: L'Hôpital's rule

Any one of the three methods earns the full 3 points.

Where marks are lost¶

Showing only injectivity in part 1. Invertible onto $\mathbb{R}$ also needs the range to be all of $\mathbb{R}\text{.}$
Writing $g'(2)=1/f'(2)\text{.}$ The derivative of $f$ is evaluated at $g(2)=1\text{,}$ not at $2\text{.}$
Plugging in $y=0$ in part 3. The quotient is $0/0\text{;}$ a limit argument is required.

Try next¶

Continuity and differentiability of $x^2\sin(A/x)$
A limit via asymptotic equivalence: more limits of type $0/0\text{.}$

Symbols: element of, infinity, real numbers

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