Question

What is the equation of the tangent line of `f(x)=14x^3-4x^2e^(3x)` at `x=-2`?

Answer 1

Find `f(-2)` and `f'(-2)` then use the tangent line formula.

The equation of the tangent is:

`y=167.56x+223,21`

Explanation:

`f(x)=14x^3-4x^2e^(3x)`

Find the derivative function:

`f'(x)=(14x^3)'-(4x^2e^(3x))'`

`f'(x)=14(x^3)'-4[(x^2)'e^(3x)+4x^2(e^(3x))']`

`f'(x)=14*3x^2-4[2xe^(3x)+4x^2*e^(3x)*(3x)']`

`f'(x)=42x^2-4[2xe^(3x)+4x^2*e^(3x)*3]`

`f'(x)=42x^2-4[2xe^(3x)+12x^2*e^(3x)]`

`f'(x)=42x^2-8xe^(3x)[1+6x]`

Finding `f(-2)`

`f(x)=14x^3-4x^2e^(3x)`

`f(-2)=14*(-2)^3-4*(-2)^2e^(3*(-2))`

`f(-2)=32e^(-6)-112`

`f(-2)=111.92`

and `f'(-2)`

`f'(x)=42x^2-8xe^(3x)[1+6x]`

`f'(-2)=42*(-2)^2-8*(-2)e^(3*(-2))[1+6*(-2)]`

`f'(-2)=168-176e^(-6)`

`f'(-2)=167.56`

Now the derivative definition:

`f'(x)=(y-f(x_0))/(x-x_0)`

If `x_0=-2`

`f'(-2)=(y-f(-2))/(x-(-2))`

`167.56=(y-111.92)/(x+2)`

`167.56(x+2)=y-111.92`

`y=167.56x+167.56*2+111.92`

`y=167.56x+223,21`

graph{14x^3-4x^2e^(3x) [-227, 254, -214.3, 26.3]}

As you can see above, the graph is increasing at a big rate for `x<0` so the big slope is actually justified.

Note: if you are not allowed to use a calculator, then you just have to carry over the `e^(-6)` all along. Keep in mind the rules for powers:

`e^(-6)=1/e^6`

`e^(-6)*e^(-6)=(e^(-6))^2=e^(-6*2)=e^(-12)`