Determine all values of x, (if any), at which the graph of the function has a horizontal tangent.
y(x) = 6x/(x-9)^2
I have these answer choices:
x=9 and x=6
x=-9
x=-9 and x=6
x= 6
The graph has no horizontal tangents.
To find the values of x at which the graph of the function has a horizontal tangent, we need to differentiate the function and find where the derivative is equal to zero.
Step 1: Find the derivative of the function with respect to x:
y'(x) = (6(x-9)^2 - 6x(2(x-9)))/((x-9)^4)
= (6(x^2 - 18x + 81) - 12x(x-9))/((x-9)^4)
= (6x^2 - 108x + 486 - 12x^2 + 108x)/((x-9)^4)
= ( -6x^2 + 486 )/((x-9)^4)
Step 2: Set the derivative equal to zero and solve for x:
-6x^2 + 486 = 0
Step 3: Solve for x:
-6x^2 = -486
x^2 = 486/6
x^2 = 81
x = ±√81
x = ± 9
So, the function has a horizontal tangent at x = 9 and x = -9.
To find the values of x at which the graph of the function has a horizontal tangent, we need to find the points where the derivative of the function is equal to zero.
1. Start by finding the derivative of the given function y(x) = 6x/(x-9)^2 using the quotient rule.
y'(x) = [6(x-9)^2 - 6x(2)(x-9)] / (x-9)^4
Simplifying this expression:
y'(x) = [6(x^2 - 18x + 81) - 12x(x-9)] / (x-9)^4
y'(x) = (6x^2 - 108x + 486 - 12x^2 + 108x) / (x-9)^4
y'(x) = (-6x^2 +486) / (x-9)^4
2. Now, we set the numerator of the derivative equal to zero since we are looking for the values of x where the derivative is equal to zero.
-6x^2 + 486 = 0
-6x^2 = -486
Divide both sides by -6 to isolate x^2:
x^2 = 81
3. Take the square root of both sides to solve for x:
x = ± √81
x = ± 9
The values of x where the graph of the function has a horizontal tangent are x = 9 and x = -9.
by quotient rule:
dy/dx = ( (x-9)^2 (6) - 6x(2)(x-9))/(x-9)^4
= 0 at a horizontal tangent
6(x-9)^2 - 12x(x-9) = 0
6(x-9)[x-9 - 2] = 0
6(x-9)(x-11) = 0
x=9 or x=11 , but x≠9 , there is a vertical asymptote at x=9
x = 11