f(x)= x^(2/3)-x^(2)

Question

this is a curve sketch. determine domain and range. find x intercepts. find y intercept. determine any vertical asymptotes. find the end behavior asymptote. consider symmetry. find the relative extrema. find the points of inflection. graph.

Answer 1

Here’s a sketch http://bit.ly/sks23cu131208wa01 that I did with GeoGebra.

From the sketch:
1. domain: All real numbers.
2. range: y <= 0.3849
3. x intercepts: -1, 0, 1
4. y intercept: 0
5. vertical asymptote: y-axis for both f and f’
6. end behavior asymptote: y = -x^2
7. symmetry: even function
8. relative extrema: (+-0.4387, 0.3849)
9. points of inflection: none b/c f’’ never zero
10. graph: see sketch

extra: f '' has a horizontal asymptote at y = -2

 

Thomas R did an excellent job of discussing the calculus. He needs to add x = 0 to item 5, though.

Answer 2

f(x)

 

Answer 3

Dear Katelun,

 

Obviously f(x) = 0 when x = 0 and x= 1.  

 

d[f(x)](dx) = (2/3)*(x^(-1/3)) - 2x

 

Point of inflection at x = (3)^(1/4)/3 and y = (2*√3)/9

 

This is a very strange function.

 

Try graphing the function using http://www.meta-calculator.com/online/

and you'll see what I mean.

Answer 4

Katelyn, hello.  There are two main ways to work on this problem.  You can use a graphing utility such as a TI-84 Calculator or you can do it entirely by hand which requires calculus.  I would use both.  You can answer some of the questions fairly easily.  

1. The y intercept is zero, since when x=0, y=0.

2. The end behavior is that of y=-x² , since as x gets very large or very small the x^2/3 term has less and less influence upon the value of y.

3. The domain is (-infinity, infinity), since there are no values of x that would cause division by zero or complex results.

4. The function is even (has y axis symmetry) , since f(-x)=f(x).

5. The x intercepts can be found by setting x^2/3-x²=0 .  Solving gives x=+1 or -1.

6. Taking the limit as x tends towards infinity or negative infinity tells us that the function has no horizontal asymptotes.

7. The fact that the function has no possibility of division by zero and its domain is all real numbers indicates that it has no vertical asymptotes either.

8. Setting the first derivative equal to zero [(2/3)x^-1/3 -2x=0] and solving gives possible x values for minimums and maximums.  We get x=(1/3)^3/4 and x=-(1/3)^3/4.  These are approximately 0.439 and  

-0.439.

9. Setting the second derivative equal to zero [(-2/9)x^-4/3-2=0] and solving we see that there is no solution.  Further investigation shows that the second derivative is always negative except at x=0 where it does not exist.  This means that the function is concave down.  Also, there are no inflection points.

10. Since the function is concave down, the function has maximums at the x values found in part 8.  We can plug these values into the function to get the maximum height.  This tells us the range which is (-infinity, sqrt3/3-(sqrt3/3)³. This is a maximum of approximately 0.385. 

11. Using this information you can sketch the function and verify that it matches what you graphed on the calculator.

I hope this helps.  Good luck.

Answer 5

             

 X^{2/3  -} X²

 

   Negative numbers can have a odd order of roots:

 

    Domain: ( -∞, + ∞)

    Range  : ( -∞, +∞ ) .