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?
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)
To determine the coordinates of the point(s) on the graph of y=3x-1/x at which the slope of the tangent is 7, you need to set the derivative equal to 7 and solve for x. Here's how you can do it:
1. Set the derivative equal to 7: y' = 3 + 1/x^2 = 7.
2. Subtract 3 from both sides of the equation: 1/x^2 = 7 - 3.
3. Simplify the equation on the right side: 1/x^2 = 4.
4. Take the reciprocal of both sides: x^2 = 1/4.
5. Take the square root of both sides: x = ±√(1/4).
6. Simplify the equation: x = ±1/2.
Now that you have the x-values, you can substitute them back into the original equation y = 3x - 1/x to find the corresponding y-values:
When x = 1/2, y = 3(1/2) - 1/(1/2) = 3/2 - 2 = -1/2.
So one of the points is (1/2, -1/2).
When x = -1/2, y = 3(-1/2) - 1/(-1/2) = -3/2 - (-2) = -3/2 + 2 = 1/2.
So the other point is (-1/2, 1/2).
Therefore, the coordinates of the points on the graph of y = 3x - 1/x at which the slope of the tangent is 7 are (1/2, -1/2) and (-1/2, 1/2).