For which values of x does the curve Y= (3x^3-3x^2+1) / (2x^2-1) , have a horizontal tangent?
Show steps or explain, i dont get it
Jaime/Mikey/Jessi/Carmen -- please use the same name for your posts.
Take the first derivative, set it equal to zero,
and solve for x
That is your x
To find the values of x for which the curve has a horizontal tangent, we need to find the derivative of the function and set it equal to zero.
First, let's find the derivative of the function. Using the quotient rule:
y = (3x^3 - 3x^2 + 1) / (2x^2 - 1)
To differentiate, we apply the quotient rule:
dy/dx = [(2x^2 - 1)(9x^2 - 6x) - (3x^3 - 3x^2 + 1)(4x)] / (2x^2 - 1)^2
Simplifying this expression gives:
dy/dx = (18x^4 - 15x^3 + 8x - 1) / (2x^2 - 1)^2
Now, to find the values of x for which the derivative dy/dx is zero, we set the numerator equal to zero:
18x^4 - 15x^3 + 8x - 1 = 0
At this point, we can either attempt to find the roots of this equation manually, which may be quite challenging, or we can use numerical methods or graphing software to determine the approximate values of x.
One way to find the approximate values of x is to graph the function or use an online graphing calculator. The x-values at which the function intersects the x-axis are the approximate values of x where the curve has horizontal tangents.
Alternatively, we can use numerical methods like the Newton-Raphson method, bisection method, or any other root-finding algorithm to obtain the approximate values of x.
Note that finding exact values for x may not be possible since the derivative equation involves a quartic polynomial (polynomial of degree 4).
Once the values of x where the derivative is zero are found, those are the values for which the curve has horizontal tangents.