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Consider the function: $$f(x)=\sqrt[3]{8x^2+4x+1}$$ 1) Find $a,b,\alpha,\beta$ such that: $$f(x)=ax^\alpha+bx^\beta+o(x^{-1/3})$$ 2) Find $A=f([0,+∞[)$ and prove that $f:[0,+∞[\rightarrow A$ is invertible.

3) Find $d, \delta$ such that $$f^{-1}(y)=dy^{\delta}+o(y^{3/2})$$

I don't understand how to use Taylor's theorem with non integer exponents in the error.

While for the point 2) I think $A=[1,+∞)$ because of intermediate value theorem since $f(x)$ is continuous, $f(0)=1$ and $\lim\limits_{x \to +∞}f(x)=+∞$. While $f:[0,+∞[\rightarrow A$ is invertible because $f'(x)>0$ $\forall x>-4$ so $f(x)$ is monotone on $[0,+∞)$ thus invertible. Thanks.

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1 Answer 1

I think i solved the point 1) by myself.

$$f(x)=(8^x2+4x+1)^{1/3}=2x^{2/3}\left(1+\frac{1}{2x}+\frac{1}{8x^2}\right)^{1/3}$$

Now for $x \rightarrow +∞$: $$\frac{1}{8x^2}=o\left(\frac{1}{x}\right)$$

so $$f(x)=2x^{2/3}\left(1+\frac{1}{2x}+o\left(\frac{1}{x}\right)\right)^{1/3}$$

and for $y \rightarrow +∞$ $$(1+y)^{1/3}=1+\frac{1}{3}y+o\left(y\right)$$

Replacing: $$f(x)=2x^{2/3}\left(1+\frac{1}{6x}+o\left(\frac{1}{x}\right)\right)=2x^{2/3}+\frac{1}{3}x^{-1/3}+o(x^{-1/3})$$

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