Determine all values of x, (if any), at which the graph of the function has a horizontal tangent.
y(x) = 6x/(x-9)^2
when i workout this problem I get this:
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
but I have these answer choices to choose from:
A. x=9 and x=6
B. x=-9
C. x=-9 and x=6
D. x= 6
E. The graph has no horizontal tangents.
is it E then
you lost an x there when factoring out the 6(x-9)
y' = -6(x+9)/(x-9)^3
y'=0 at x = -9
No, it is not choice E. To determine the values of x at which the graph of the function has a horizontal tangent, we need to find the values of x for which the derivative of the function is equal to zero.
In this case, the derivative is:
dy/dx = ( (x-9)^2 * (6) - 6x * (2)(x-9) ) / (x-9)^4
To find the values of x when dy/dx = 0, we need to solve the equation:
6(x-9)^2 - 12x(x-9) = 0
Simplifying the equation, we have:
6(x-9)(x-9-2) = 0
6(x-9)(x-11) = 0
So, x could be either 9 or 11. However, since the original function has a vertical asymptote at x=9, we can eliminate x=9 as a valid value.
Therefore, the only value of x at which the graph of the function has a horizontal tangent is x=11.
Thus, the correct answer is:
D. x=11