Graph of a function in the neighbourhood of $x=0$

Question

I'm new in Calculus II. Given the function $f(x)=(x^2(e^x−1))^{\frac{1}{5}}$, I would like to determine a qualitative graph of the function in the neighbourhood of x=0. I know that there might be an inflection point in x=0, but I'm stuck as I don't really know how to proceed. Should I calculate the second derivative in $x=0$?

Answer 1

This problem seems designed for using power series expansions (i.e. Taylor series), which is usually covered towards the end of U.S. calculus 2 courses, but since you didn't say what your Calculus II consists of (or even say what you've covered so far), I'm going to take the power series approach.

In what follows I'll show that

$$ x^{2/5}(e^x - 1)^{1/5} \;\; = \;\; x^{3/5} \; + \; \frac{1}{10}x^{8/5} \; + \; \frac{1}{75}x^{13/5} \; + \; \cdots $$

I’ll avoid taking derivatives by using basic algebraic manipulation along with the expansions of $e^u$ and $(1+u)^{1/5}:$

$$e^u \;\; = \;\; 1 \; + \; u \; + \; \frac{1}{2}u^2 \; + \; \frac{1}{6}u^3 \; + \; \frac{1}{24}u^4 \; + \; \cdots $$

$$ (1+u)^{1/5} \;\; = \;\; 1 \; + \; \frac{1}{5}u \; - \; \frac{2}{25}u^2 \; + \; \frac{6}{125}u^3 \; - \; \cdots $$

The expansion for $(1+u)^{1/5}$ is the binomial expansion for the exponent $\frac{1}{5}.$ See this answer, using $t=\frac{1}{5}.$

In what follows I’ll carry out the computations far enough to obtain the first three terms of the expansion.

$$ x^{2/5}(e^x - 1)^{1/5} \;\; = \;\; x^{2/5}\cdot \left[ \left( 1 + x + \frac{1}{2}x^2 + \frac{1}{6}x^3 + \cdots \right) \; - \; 1 \right]^{1/5} $$

$$ = \;\; x^{2/5}\cdot \left[x \; + \; \frac{1}{2}x^2 \; + \; \frac{1}{6}x^3 \; + \; \cdots \right]^{1/5} $$

$$ = \;\; x^{2/5}\cdot \left[ x\left(1 \; + \; \frac{1}{2}x \; + \; \frac{1}{6}x^2 \; + \; \cdots \right)\right]^{1/5} $$

$$ = \;\; x^{2/5}\cdot x^{1/5} \cdot \left[1 \; + \; \frac{1}{2}x \; + \; \frac{1}{6}x^2 \; + \; \cdots \right]^{1/5} $$

$$ = \;\; x^{3/5}\cdot \left[1 \; + \; \left(\frac{1}{2}x \; + \; \frac{1}{6}x^2 \; + \; \cdots\right) \right]^{1/5} $$

$$ = \;\; x^{3/5}\cdot \left[1 \; + \; u \right]^{1/5}, $$

where $u \; = \; \left(\frac{1}{2}x + \frac{1}{6}x^2 + \cdots \right).$

Thus, we have

$$ x^{2/5}(e^x - 1)^{1/5} \;\; = \;\; x^{3/5}\cdot \left[1 \; + \; u \right]^{1/5} $$

$$= \;\; x^{3/5} \cdot \left[ 1 \; + \; \frac{1}{5}u \; - \; \frac{2}{25}u^2 \; + \; \frac{6}{125}u^3 \; - \; \cdots \right] $$

$$= \;\; x^{3/5} \cdot \left[ 1 + \frac{1}{5}\left(\frac{1}{2}x + \frac{1}{6}x^2 + \cdots \right) - \frac{2}{25}\left(\frac{1}{2}x + \frac{1}{6}x^2 + \cdots \right)^2 + \frac{6}{125}\left(\frac{1}{2}x + \frac{1}{6}x^2 + \cdots \right)^3 - \cdots \right] $$

$$= \;\; x^{3/5} \cdot \left[ 1 + \frac{1}{5}x\left(\frac{1}{2} + \frac{1}{6}x + \cdots \right) - \frac{2}{25}x^2\left(\frac{1}{2} + \frac{1}{6}x + \cdots \right)^2 + \frac{6}{125}x^3\left(\frac{1}{2} + \frac{1}{6}x + \cdots \right)^3 - \cdots \right] $$

To obtain the correct first three terms, we need to obtain the correct first three terms inside the square brackets (i.e. up through quadratic in $x).$ In everything that follows, I'll use $\;+ \; \cdots \;$ at the point where we do not need to continue any further, given that all we want are the correct terms up through quadratic in $x.$

$$ \frac{1}{5}x\left(\frac{1}{2} + \frac{1}{6}x + \cdots \right) \;\; = \;\; 0 \; + \;\frac{1}{10}x \; + \; \frac{1}{30}x^2 \; + \; \cdots $$

$$ \frac{2}{25}x^2\left(\frac{1}{2} + \frac{1}{6}x + \cdots \right)^2 \;\; = \;\; \frac{2}{25}x^2 \cdot \left(\frac{1}{2}\right)^2 \; + \; \cdots \;\; = \;\; 0 \; + \; 0x \; + \; \frac{1}{50}x^2 \; + \; \cdots $$

$$ \frac{6}{125}x^3\left(\frac{1}{2} + \frac{1}{6}x + \cdots \right)^3 \;\; = \;\; 0 \;+ \; 0x \; + \; 0x^2 \; + \; \cdots $$

Using these, we get

$$= \;\; x^{3/5} \cdot \left[ 1 + \left(0 + \frac{1}{10}x + \frac{1}{30}x^2 + \cdots\right) - \left(0 + 0x + \frac{1}{50}x^2 + \cdots\right) + \left(0 + 0x + 0x^2 + \cdots\right) - \cdots \right] $$

$$= \;\; x^{3/5} \cdot \left[ 1 \; + \; \frac{1}{10}x \; + \; \frac{1}{30}x^2 \; - \; \frac{1}{50}x^2 \; + \; \cdots \right] $$

$$= \;\; x^{3/5} \cdot \left[ 1 \; + \; \frac{1}{10}x \; + \; \frac{1}{75}x^2 \; + \; \cdots \right] $$

$$ = \;\; x^{3/5} \; + \; \frac{1}{10}x^{8/5} \; + \; \frac{1}{75}x^{13/5} \; + \; \cdots $$

The expansion above agrees with what WolframAlpha gives.

See here for a graph of $y = x^{2/5}(e^x - 1)^{1/5}$ and $y = x^{3/5}$ for $0 < x < 4$ on the same axes.

See here for a graph of $y = x^{2/5}(e^x - 1)^{1/5}$ and $y = x^{3/5} + \frac{1}{10}x^{8/5}$ for $0 < x < 4$ on the same axes.

See here for a graph of $y = x^{2/5}(e^x - 1)^{1/5}$ and $y = x^{3/5} + \frac{1}{10}x^{8/5} + \frac{1}{75}x^{13/5}$ for $0 < x < 4$ on the same axes.