This Question is on CALCULUS:
Find the point on the curve f(x) = x2 (X Squared) - 6x + 11 where the tangent is parallel to the x-axis?
Thank you :)
is it h.l or o.l
Here's the solution:
The tangent of the curve parallel to the x-axis will have a slope of 0.
Then we find the derivative of f(x) = x^2 - 6x + 11 to find the slope of any tangent.
dy/dx = 2x - 6
If slope = 0,
2x - 6 = 0
2x = 6
x = 3
Then we find the corresponding y-value for x = 3
y = (3)^2 - 6(3) + 11 = 9 - 18 + 11 = 2
Therefore the point is (3,2)