## University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson

# Chapter 3 - Section 3.1 - Tangents and the Derivative at a Point - Exercises - Page 119: 46

#### Answer

The graph of the given function has 2 vertical tangents at $x=0$ and $x=1$.

#### Work Step by Step

$$y=f(x)=x^{1/3}+(x-1)^{1/3}$$ a) The graph is enclosed below. From the introduction, we can guess that the graph appears to have 2 vertical tangents, one at $x=0$ and the other at $x=1$. We will test this in part b). b) The graph of $f(x)$ has a vertical tangent at $x$ when 1) $f$ is continuous at $x$ 2) $$f′(x)=\lim_{h\to0}\frac{f(x+h)−f(x)}{h}=\pm\infty$$ First, for $f'(0)$: $$f′(0)=\lim_{h\to0}\frac{f(h)−f(0)}{h}=\lim_{h\to0}\frac{h^{1/3}+(h-1)^{1/3}-\Big(0^{1/3}+(0-1)^{1/3}\Big)}{h}$$ $$f'(0)=\lim_{h\to0}\frac{h^{1/3}+(h-1)^{1/3}-(0-1)}{h}=\lim_{h\to0}\frac{h^{1/3}+(h-1)^{1/3}+1}{h}$$ $$f'(0)=\lim_{h\to0}\frac{(h-1)^{1/3}+1}{h}+\lim_{h\to0}\frac{h^{1/3}}{h}=A+B$$ $$A=\lim_{h\to0}\frac{(h-1)^{1/3}+1}{h}$$ Multiply both numerator and denominator by $(h-1)^{2/3}-(h-1)^{1/3}+1$: $$A=\lim_{h\to0}\frac{\Big((h-1)^{1/3}+1\Big)\Big((h-1)^{2/3}-(h-1)^{1/3}+1\Big)}{h\Big((h-1)^{2/3}-(h-1)^{1/3}+1\Big)}$$ $$A=\lim_{h\to0}\frac{h-1+1}{h\Big((h-1)^{2/3}-(h-1)^{1/3}+1\Big)}$$ $$A=\lim_{h\to0}\frac{h}{h\Big((h-1)^{2/3}-(h-1)^{1/3}+1\Big)}$$ $$A=\lim_{h\to0}\frac{1}{(h-1)^{2/3}-(h-1)^{1/3}+1}$$ $$A=\frac{1}{(0-1)^{2/3}-(0-1)^{1/3}+1}=\frac{1}{1-(-1)+1}=\frac{1}{3}$$ $$B=\lim_{h\to0}\frac{h^{1/3}}{h}=\lim_{h\to0}\frac{1}{h^{2/3}}$$ - As $h\to0^+$ or $h\to0^-$, $h^{2/3}$ always approaches $0^+$, so $\lim_{h\to0}\frac{1}{h^{2/3}}=\infty$. In conclusion, $$f'(0)=A+B=\frac{1}{3}+\infty=\infty$$ This means $f$ has a vertical tangent at $x=0$ as predicted. Second, for $f'(1)$: $$f′(1)=\lim_{h\to0}\frac{f(h+1)−f(1)}{h}$$ $$f'(1)=\lim_{h\to0}\frac{\Big((h+1)^{1/3}+(h+1-1)^{1/3}\Big)-\Big(1^{1/3}+(1-1)^{1/3}\Big)}{h}$$ $$f'(1)=\lim_{h\to0}\frac{\Big((h+1)^{1/3}+h^{1/3}\Big)-(1+0)}{h}$$ $$f'(1)=\lim_{h\to0}\frac{(h+1)^{1/3}+h^{1/3}-1}{h}=\lim_{h\to0}\frac{(h+1)^{1/3}-1}{h}+\lim_{h\to0}\frac{h^{1/3}}{h}$$ $$f'(1)=A+B$$ $$A=\lim_{h\to0}\frac{(h+1)^{1/3}-1}{h}$$ Multiply both numerator and denominator by $(h+1)^{2/3}+(h+1)^{1/3}+1$: $$A=\lim_{h\to0}\frac{\Big((h+1)^{1/3}-1\Big)\Big((h+1)^{2/3}+(h+1)^{1/3}+1\Big)}{h\Big((h+1)^{2/3}+(h+1)^{1/3}+1\Big)}$$ $$A=\lim_{h\to0}\frac{(h+1)-1}{h\Big((h+1)^{2/3}+(h+1)^{1/3}+1\Big)}$$ $$A=\lim_{h\to0}\frac{h}{h\Big((h+1)^{2/3}+(h+1)^{1/3}+1\Big)}$$ $$A=\lim_{h\to0}\frac{1}{(h+1)^{2/3}+(h+1)^{1/3}+1}$$ $$A=\frac{1}{(0+1)^{2/3}+(0+1)^{1/3}+1}=\frac{1}{3}$$ $$B=\lim_{h\to0}\frac{h^{1/3}}{h}=\lim_{h\to0}\frac{1}{h^{2/3}}$$ - As $h\to0^+$ or $h\to0^-$, $h^{2/3}$ always approaches $0^+$, so $\lim_{h\to0}\frac{1}{h^{2/3}}=\infty$. In conclusion, $$f'(1)=A+B=\frac{1}{3}+\infty=\infty$$ This means $f$ has a vertical tangent at $x=1$ as predicted.

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