So here is the question:
Find and equation of the tangent line to the graph:
4)f(x)=e^(2x) passing through the point (0,0)
I find derivative being f'(x) =2e^(2x)
Then m = f'(0) = 2
put into point slope formula:
(y-y1)=m(x-x1)
y-0 = 2(x-0)
which comes out to
y=2x
0=2x ; so the new x coordinate is x= 0
new y is y= 2(0) = 0
new m = 2e^(2(0)) = 2e^(0) = 2(1) = 2
therefore final function is y = 2x
Is this right?
As you can see from the graphs at
http://www.wolframalpha.com/input/?i=plot+y%3De^%282x%29%2C+y%3D2x
you missed the mark. So, let's see what happened. At the point (x,y) on the curve, the slope is 2e^(2x).
A line with slope 2e^(2x) through the origin:
e^(2x)/x = 2e^(2x)
x = 1/2
So, the line is tangent to the curve at (1/2,e), with slope 2e. See
http://www.wolframalpha.com/input/?i=plot+y%3De^%282x%29%2C+y%3D2e*x
Thanks!
Yes, your solution is correct. The tangent line equation to the graph of the function f(x) = e^(2x) passing through the point (0,0) is indeed y = 2x.
To arrive at this solution, you correctly found the derivative of the function f(x) = e^(2x) to be f'(x) = 2e^(2x). Then, you evaluated f'(0) to find the slope of the tangent line as m = 2.
Using the point-slope form of a linear equation, you substituted the slope and the given point (0,0) into the equation y - y1 = m(x - x1) to obtain y - 0 = 2(x - 0), which simplifies to y = 2x.
To double-check your solution, you also recalculated the x and y coordinates using the equation y = 2x. Plugging in x = 0, you correctly obtained y = 2(0) = 0. Additionally, you confirmed the slope of the tangent line to be 2 by evaluating f'(0) and finding that 2e^(2(0)) = 2(1) = 2.
Hence, your final function y = 2x is indeed the correct equation for the tangent line to the graph of f(x) = e^(2x) passing through the point (0,0). Well done!