Question

Can someone help me? thankss
Find the equation of the the tangent line to the curve:
y=1 + xe^2x

Answer 1

At which point?

Answer 2

x=0

Answer 3

Take the derivative of the function, evaluate at x=0; that will be the slope of your line (m). Use the point/slope form of the equation of the line by using that value of m and (0,f(0)).

Answer 4

Alright thank you, but another question. How would you solve for something like this:

Answer 5

No such (real) value of x exists. Exponential functions are strictly positive.

Answer 6

Okay but in general, how woul you solve an equation with e?

Answer 7

use the product rule to find the derivative

u = x du/dx = 1

v = e^2x dv/dx = 2e^2x

then

\[dy/dx = 2xe^{2x} + e^{2x} = e^{2x}( 2x + 1)\]

now substitute x = 0 to find the gradient of the tangent

\[m = 2^{2\times0}(2\times0 + 1) = 1\]

the tangent has a gradient of 1 at x = 0

the point on the curve when x = 0 is y = 1

so you have a gradient of m = 1 and point (0,1)

then the tangent is

y = 1 = 1(x - 1)

y = x + 1 is the equation of the tangent at x = 0

Answer 8

thank you all for your answers :D