Quick question help please (1 Viewer)

[tonysoprano]

Hey everyone,

Could anybody please help solve this question?

The line y=mx is a tangent to the curve y=e^3x. Find m.

The answer is 3e.

Thanks!

 

[photastic]

Firstly, the y values of the line and curve will intercept. y = mx = e^3x.
Now the gradients of the line and the tangent will intercept, m = 3e^3x.
Combine the two equations, m = 3mx, x = 1/3.
Therefore since mx = e^3x, now when x = 1/3,
m(1/3) =e^3(1/3)
m/3 = e^1
m = 3e

 

[tonysoprano]

Firstly, the y values of the line and curve will intercept. y = mx = e^3x.
Now the gradients of the line and the tangent will intercept, m = 3e^3x.
Combine the two equations, m = 3mx, x = 1/3.
Therefore since mx = e^3x, now when x = 1/3,
m(1/3) =e^3(1/3)
m/3 = e^1
m = 3e

Oh yup, was lost for a second when reading that, but I got it. Thanks so much!