3 + (1/x^2) = 7
1/x^2 = 4
x^2 = (1/4)
x = .5 or -.5
when x = .5
y = 1.5 - 2 = -.5 so (.5, -.5)
when x = -.5
y = -1.5 +2 = .5 so (-.5, .5)
Determine the coordinates of the point(s) on the graph of y=3x- 1/x at which the slope of the tangent is 7.
So the derivative of y=3x- 1/x is
y'=3 + 1/x^2
What do I do next?
1 answer