Question

At what point on the graph of \(y=2 e^{x}-1\) is the tangent line perpendicular to the line \(y=-3 x+2 ?\)

Short answer

The tangent line at the point \(( ln(1/6), -1/3 )\) on the graph of \(y=2 e^{x}-1\) is perpendicular to the given line \(y=-3x+2\).

Step-by-step solution

Find the derivative of the function

The derivative of a function gives the slope of the function at any point x. The derivative of \(y=2 e^{x}-1\) is \(y'=2 e^{x}\) .

Find the slope of the given line

The equation of the given line is in the slope-intercept form \(y = mx + c\), where m is the slope. So, the slope of \(y=-3x+2\) is -3.

Set the product of the slopes to -1

Two lines are perpendicular if the product of their slopes is -1. So, set \(y'*(-3) = -1\), then solve for x. This gives \((-3)*(2e^{x}) = -1\). Solving this equation gives \(x= ln(1/6)\).

Find the y-coordinate

Substitute \(x= ln(1/6)\) into the equation \(y=2e^{x}-1\) to get \(y = 2e^{ln(1/6)} - 1 = 2*1/6 - 1 = -1/3\)