Question

a continuous curve y = f(x) has a vertical tangent line at the point where x = x₀ if lim(h→0) (f(x₀ + h) - f(x₀))/h = ∞ or -∞.
a. graph the curve y = 2x^(3/5). where does the graph appear to have vertical tangent lines?
b. confirm your findings in part (a) with limit calculations.
a. select the correct graph below. the graphs are shown in a -10,10 by -10,10 viewing window.

Explanation:

Step1: Recall the function form

The function is $y = 2x^{\frac{3}{5}}$.

Step2: Analyze the graph visually

The graph of $y = 2x^{\frac{3}{5}}$ has a cusp - like behavior at $x = 0$. It appears to have a vertical tangent at $x=0$.

Step3: Calculate the limit

First, find $f(x + h)-f(x)$ for $f(x)=2x^{\frac{3}{5}}$.
\[

$$\begin{align*} f(x + h)&=2(x + h)^{\frac{3}{5}}\\ f(x + h)-f(x)&=2(x + h)^{\frac{3}{5}}-2x^{\frac{3}{5}} \end{align*}$$

\]
Then, find $\lim_{h
ightarrow0}\frac{f(x + h)-f(x)}{h}=\lim_{h
ightarrow0}\frac{2(x + h)^{\frac{3}{5}}-2x^{\frac{3}{5}}}{h}$. Let $x = 0$, then we have $\lim_{h
ightarrow0}\frac{2(0 + h)^{\frac{3}{5}}-2(0)^{\frac{3}{5}}}{h}=\lim_{h
ightarrow0}\frac{2h^{\frac{3}{5}}}{h}=\lim_{h
ightarrow0}2h^{-\frac{2}{5}}=\infty$ as $h
ightarrow0^{+}$ and $-\infty$ as $h
ightarrow0^{-}$.

a. The graph of $y = 2x^{\frac{3}{5}}$ has a vertical tangent at $x = 0$. Without seeing the actual graphs A - D, we know the general shape of $y = 2x^{\frac{3}{5}}$ which is symmetric about the origin and has a vertical tangent at $x = 0$.
b. We have confirmed with limit calculations that at $x = 0$, $\lim_{h
ightarrow0}\frac{f(0 + h)-f(0)}{h}=\lim_{h
ightarrow0}\frac{2h^{\frac{3}{5}}}{h}=\lim_{h
ightarrow0}2h^{-\frac{2}{5}}$ is either $\infty$ or $-\infty$ depending on the sign of $h$ approaching $0$.

Answer:

a. The graph appears to have a vertical tangent line at $x = 0$.
b. We have confirmed through limit calculations that the function $y = 2x^{\frac{3}{5}}$ has a vertical tangent line at $x = 0$ since $\lim_{h
ightarrow0}\frac{f(0 + h)-f(0)}{h}=\lim_{h
ightarrow0}\frac{2h^{\frac{3}{5}}}{h}=\lim_{h
ightarrow0}2h^{-\frac{2}{5}}=\pm\infty$ as $h
ightarrow0^{\pm}$.