calculus Question

Question

Use the function f(x) = x^2-4√x to answer the following questions.

A) Is the graph symmetric about the x or y axis? Justify your answer
B) Find the asymptotes if any
C) Where does the function increase and decrease?
D) Determine the maximum and minimum values

Help!?

Answer 1

A)

To test for Symmetry about the x-axis, replace the y with (-y) and simplify. If it is symmetric, the equation will be the same as the original, since symmetric means the graph will be the same above and below the y axis. 

To test for Symmetry about the y-axis, replace the x with (-x) and simplify. The equation will the the same as the original if it is symmetric. 

 

x-axis

Original: y =x²-4√x

(-y) = x²-4√x

This is not the same as the original, so it is not symmetric about the x-axis

 

y-axis

y=(-x)²-4√(-x)=x²-4√(-x)

This is not the same as the original inside of the square room, also the domain changes. 

 

Therefore, there is no symmetry.

 

B)

Asymptotes are found using the derivative. 

Horizontal Asymptotes (HA) are where the derivative equals 0 (No increase).

Vertical asymptotes (VA) are where the derivative is undefined (Increasing at a very fast rate)

 

The derivative is found to be y' = 2x - 2/(√x)

Set this equal to zero and solve for x to find HA.

2x - 2/(√x) = 0

Solving this equation for x, you get that x=1, so there is a HA at x=1. Plug x=1 into the original equation if you want to know the value of the y-coordinate. This may not be necessary, but it depends how your professor wants the final answer. Plugging in x=1 into the original gives the equation of the HA as y=-3

 

To find the VAs, determine where the derivative will be undefined. y' = 2x - 2/(√x) This will occur at x=0 since you cannot divide by 0.

 

 

C) The function is increasing when the derivative is positive, and decreasing when the derivative is negative.

You can determine these intervals by setting the derivative equal to 0, which we have previously done. 2x - 2/(√x) = 0 (x=1)

Now analyze the left and right side of the derivative at x=1

If we plug in any number in the domain that is to the left, for example x=1/4, we see that derivative produces a negative number, which means the original function is decreasing

If we analyze the right side by plugging in x=4, the derivative produces a positive number, so the original function is increasing. 

Therefore, the function is decreasing from x: [0, 1) and increasing from x: (1, infinity)

 

D)

The maximum and minimum values can be found by analyzing the location of the Horizontal asymptote, since this is where the graph changes directions.

Setting the derivative equal to 0 (2x - 2/(√x) = 0) gives us x=1. This is sometimes called a critical point. Plug in x=1 into the ORIGINAL equation to determine the value, and if it is a MINIMUM or MAXIMUM

Plugging in x=1 into y =x²-4√x gives y=-3. Therefore, this is either a Maximum or Minimum. To determine if it is a maximum or minimum, you can analyze if the function is increasing or decreasing to the left and right of x=1 by using the information from part C. Since the function is decreasing to the left of x=1, and increasing to the right of x=1, this must be a MINIMUM.

Another way to check if it is a Minimum or maximum is by plugging in a value to the left or right of x=1 into the original equation and analyze if that number is smaller or larger than y=-3. In this case, plugging in any number that is in the domain which is to the left or right of x=1 into the original equation produces a number that is larger than y=-3, so y=-3 is the smallest possible value, which means that it is a minimum. 

Answer 2

I did not mean to put my solution in the comments.

 

 

A) If f(x) = f(-x), then symmetrical to y-axis.
    If f(x) = -f(x), then symmetrical to x-axis.
 
 
B)
 
We cannot have any negative values in the domain since there is a square-root. But since we don't have a rational function, there is not vertical asymptote.  Just a discontinuity at negative numbers.
 
Restrictions are all real negatives numbers.
 
If you evaluate the limit as x-->∞ , you will find your horizontal asymptote.  If the limit is not a constant number, then there is no horizontal asymptote.
 
 
C and D)
 
To find where function is increasing, set the derivative of f(x) equal to zero.  Then solve for x.
 
f(x) = x² - 4x^1/2
 
f'(x) = 2x - 2x^-1/2
 
f'(x) = 2x - (2 / √x)
 
2x - (2 / √x) = 0
 
2x^3/2 - 2 = 0
 
2(x^3/2 - 1) = 0
 
x^3/2 - 1 = 0
 
x^3/2 = 1
 
x = 1
 
Now just evaluate f'(0) and f'(2).
 
If f'(0) is positive and f'(2) is negative, you have a maximum at x=1.
If f'(0) is negative and f'(2) is positive, you have a minimum at x=1.
 
Based on this derivative test, you can tell where the function is increasing and decreasing.