iRaid said:Homework Statement
Find an equation for a line that is tangent to the grpah of y=ex and goes through the origin.
Homework Equations
The Attempt at a Solution
y'=ex
That's about all I can think of. I don't know how to make the tangent line go through the origin. Can someone lead me in the right direction?
Ray Vickson said:At any point (x,y) on the graph, how would you compute the slope of the tangent line?
RGV
The formula for the tangent line to y=e^x at the origin is y=x.
The slope of the tangent line to y=e^x at the origin is equal to the derivative of the function at the point of interest. In this case, the derivative of y=e^x is e^x, so the slope at the origin is e^0, which equals 1.
The tangent line to y=e^x at the origin represents the instantaneous rate of change of the function at that point. It also serves as a linear approximation of the function near the origin.
Yes, since the equation for the tangent line is y=x, any point on this line can be substituted into the equation for y to find the corresponding value of x, and vice versa.
The equation for the tangent line can be derived using the slope-intercept form of a line: y=mx+b. By setting x=0 and y=e^x, we can solve for the y-intercept, which is equal to e^0=1. Then, using the slope calculated in question 2, we can substitute the slope and the y-intercept into the slope-intercept form to get y=x.