f(x)= (100(x-5))/(x-3)^2

Question

can you help me find the intervals it is increasing and decreasing 

also where i would find any asymptotes.

also where it is concave up or concave down

 

THANK YOU

Answer 1

f(x) = (100x - 500) / (x - 3)²

 

To find where the graph is increasing and decreasing, we set the derivative of f(x) = 0; since the maximum and minimum has a tangent line whose slope is zero, and the derivative is the slope of the tangent line. This is called the first derivative test. 

 

Use the quotient rule to find the derivative.

 

f'(x) = 0

 

[100(x - 3)² - 2(x - 3)(100x - 500)] / (x - 3)⁴ = 0

 

Now we solve for x.  The x values will be our critical points, which are the locations of the maximum or minimum.

Simplify the left side of the equation.

 

Factor out (x - 3). 

 

(x - 3)(100(x - 3) - 2(100x - 500)) / (x - 3)⁴ = 0

 

[100x - 300 - 200x + 1000] / (x - 3)³ = 0

 

(-100x + 700) / (x - 3)³ = 0

 

 

Set the numerator to zero.

 

-100x + 700 = 0

-100x = -700

       x = 7

 

Our critical point is x = 7.  Next is to perform the test point to find where the graph decrease and increase.  Since the maximum or minimum is at x = 7, let use x = 6 and x = 8.  Evaluate these points into the simplified form of the derivative.

 

 

f'(6) = (-100(6) + 700) / (6 - 3)³

       = (-600 + 700) / 3³

       = 100 / 27

 

f'(8) = (-100(8) + 700) / (8 - 3)³

       = (-800 + 700) / 5³

       = -100 / 125

 

When x= 6, the derivative is positive  When x=8, the derivative is negative.  We will have a maximum. 

 

Interval of increase is (-∞, 7).

Interval of decrease is (7, ∞).

 

A maximum indicates downward concavity.  Knowing this will save us time from performing the second derivative test, which is used to find points of infection and the shape of the graph.

 

Vertical Asymptotes

 

Set the denominator of the original function equal to zero.

 

(x - 3)² = 0

x - 3 = 0

x = 3

 

The vertical asymptote is x = 3.

 

Horizontal Asymptotes

 

We find the limit of the original function as x approaches infinity from both directions

 

lim f(x)

x-->∞

 

When we substitute ∞ into the function, we end up with ∞/∞.  This is indeterminate, so we must use L'Hospital Rule.  We take the derivative of the numerator divided by derivative of denominator.  We keep doing this until the limit is no longer indeterminate.

 

lim  100 / (2(x - 3))  = 0

x-->∞

 

The horizontal asymptote is y = 0.