Find the points on the graph of the following function where the slope of the tangent line is equal to -5/2.
f(x) = x+5/x-5
(x, y) = (smaller x-value)
(x, y) = (larger x-value)
To find the points on the graph where the slope of the tangent line is equal to -5/2 for the given function f(x) = (x+5)/(x-5), you can follow these steps:
Step 1: Take the derivative of the function f(x) to find the slope of the tangent line at any point on the graph.
The derivative of f(x) = (x+5)/(x-5) can be found using the quotient rule:
f'(x) = [(x-5)(1) - (x+5)(1)] / (x-5)^2
Simplifying further:
f'(x) = (-10) / (x-5)^2
Step 2: Find the x-values where the slope of the tangent line is -5/2.
To do this, set f'(x) = -5/2 and solve for x:
(-10) / (x-5)^2 = -5/2
Cross-multiplying:
-20 = -5(x-5)^2
Dividing both sides by -5 and taking the square root:
4 = sqrt((x-5)^2)
Squaring both sides:
16 = (x-5)^2
Step 3: Solve the equation to find the x-values.
Taking the square root of both sides:
x - 5 = ±4
Adding 5 to both sides:
x = 5 ± 4
Step 4: Find the y-values corresponding to the x-values found.
Plug in the x-values into the original function f(x) = (x+5)/(x-5):
For x = 5 + 4 = 9:
f(9) = (9 + 5) / (9 - 5) = 14 / 4 = 7/2
So, one of the points is (9, 7/2).
For x = 5 - 4 = 1:
f(1) = (1 + 5) / (1 - 5) = 6 / -4 = -3/2
So, another point is (1, -3/2).
Therefore, the points on the graph of f(x) where the slope of the tangent line is equal to -5/2 are:
(x, y) = (9, 7/2) (smaller x-value)
(x, y) = (1, -3/2) (larger x-value)