Question #104424
Trace the curve y =³√ x −1 , and state all the properties you use to trace it.
1
Expert's answer
2020-03-04T17:35:52-0500

y=f(x)=x31y=f(x)=\sqrt[3]{x} - 1

1) Domain

Domain is all real numbers.


2) Symmetry

f(x)=x31=x31f(x) or f(x)f(-x) = \sqrt[3]{-x}-1 = -\sqrt[3]{x}-1 \neq -f(x) \text{ or } f(x)

The function is neither even nor odd.


3) Intercepts  on the axes

x-intercept:

y=0x31=0x=1y=0 \\ \sqrt[3]{x} - 1 =0\\ x=1

y-intercept:

x=0y=031=1x=0\\ y=\sqrt[3]{0}-1=-1


4) Asymptotes

Let y=kx+by=kx+b is asymptote.

k=limxf(x)x=limxx31x=0k=\lim\limits_{x\to\infty}{\frac{f(x)}{x}}=\lim\limits_{x\to\infty}{\frac{\sqrt[3]{x}-1}{x}}=0

b=limx(f(x)kx)=limxx31=b=\lim\limits_{x\to\infty}{(f(x)-kx)}=\lim\limits_{x\to\infty}{\sqrt[3]{x}-1}=\infty

No asymptotes.


5) Monotonicity

y=13x23y'=\frac{1}{3\sqrt[3]{x^2}}

The first derivative does not exist when x=0.x=0. If x>0,x>0, then y>0.y'>0. If x<0,x<0, then y>0.y'>0. Therefore, y=f(x)y=f(x) is monotonically increasing and x=0x=0 is the critical point.


6) Convexity

y=29xx23y'' = -\frac{2}{9x\sqrt[3]{x^2}}

The second derivative does not exist when x=0.x=0. If x>0,x>0, then y<0.y''<0. If x<0,x<0, then y>0.y''>0. Therefore, when x<0x<0 the function is convex and when x>0x>0 the function is concave.








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