Find the line tangent to

Question

f(x)=2e^x at x=3?

Answer 1

f(x) = 2e^x

Apply derivative each side.

f'(x) = 2e^x

If x = 3 then f'(3) = 2e³

Here slope is 2e³

Now you need a point, we know it touches at x = 3, so we want (3, f(3)) = (3, 2e³)

The straight-line equations is called the "point-slope" form. For this one, they give you a point (x₁, y₁) and a slope m, and have you plug it into this formula: y - y₁ = m(x - x₁)

y - 2e³ = 2e³(x - 3)

y - 2e³ = 2xe³ - 6e³

Add 2e³ to each side.

y = 2xe³ - 4e³

The tangent line equation is y = 2xe³ - 4e³.