Question

What is the equation of the line tangent to `f(x)= x^3e^x` at `x=2`?

Answer 1

`y-8e^2=20e^2(x-2)`

Explanation:

To find the equation of the tangent line, we need a point and a slope, then we can put it into point-slope form.

Point: `(2,f(2))`
`f(2)=(2)^3e^(2)=8e^2`
`(2,8e^2)`

Slope: `f'(2)`
`f(x)=(x^3)(e^x)`
`f'(x)=(x^3)(e^x)+(3x^2)(e^x)`
`f'(x)=(x^2)(e^x)(x+3)`

`f'(2)=((2)^2)(e^2)(2+3)`
`=(4)(e^2)(5)`
`=20e^2`

Point-slope form:`y-y_1=m(x-x_1)`
`y-8e^2=20e^2(x-2)`