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

Mar 14, 2016

Find $f \left(- 2\right)$ and $f ' \left(- 2\right)$ then use the tangent line formula.

The equation of the tangent is:

$y = 167.56 x + 223 , 21$

#### Explanation:

$f \left(x\right) = 14 {x}^{3} - 4 {x}^{2} {e}^{3 x}$

Find the derivative function:

$f ' \left(x\right) = \left(14 {x}^{3}\right) ' - \left(4 {x}^{2} {e}^{3 x}\right) '$

$f ' \left(x\right) = 14 \left({x}^{3}\right) ' - 4 \left[\left({x}^{2}\right) ' {e}^{3 x} + 4 {x}^{2} \left({e}^{3 x}\right) '\right]$

$f ' \left(x\right) = 14 \cdot 3 {x}^{2} - 4 \left[2 x {e}^{3 x} + 4 {x}^{2} \cdot {e}^{3 x} \cdot \left(3 x\right) '\right]$

$f ' \left(x\right) = 42 {x}^{2} - 4 \left[2 x {e}^{3 x} + 4 {x}^{2} \cdot {e}^{3 x} \cdot 3\right]$

$f ' \left(x\right) = 42 {x}^{2} - 4 \left[2 x {e}^{3 x} + 12 {x}^{2} \cdot {e}^{3 x}\right]$

$f ' \left(x\right) = 42 {x}^{2} - 8 x {e}^{3 x} \left[1 + 6 x\right]$

Finding $f \left(- 2\right)$

$f \left(x\right) = 14 {x}^{3} - 4 {x}^{2} {e}^{3 x}$

$f \left(- 2\right) = 14 \cdot {\left(- 2\right)}^{3} - 4 \cdot {\left(- 2\right)}^{2} {e}^{3 \cdot \left(- 2\right)}$

$f \left(- 2\right) = 32 {e}^{- 6} - 112$

$f \left(- 2\right) = 111.92$

and $f ' \left(- 2\right)$

$f ' \left(x\right) = 42 {x}^{2} - 8 x {e}^{3 x} \left[1 + 6 x\right]$

$f ' \left(- 2\right) = 42 \cdot {\left(- 2\right)}^{2} - 8 \cdot \left(- 2\right) {e}^{3 \cdot \left(- 2\right)} \left[1 + 6 \cdot \left(- 2\right)\right]$

$f ' \left(- 2\right) = 168 - 176 {e}^{- 6}$

$f ' \left(- 2\right) = 167.56$

Now the derivative definition:

$f ' \left(x\right) = \frac{y - f \left({x}_{0}\right)}{x - {x}_{0}}$

If ${x}_{0} = - 2$

$f ' \left(- 2\right) = \frac{y - f \left(- 2\right)}{x - \left(- 2\right)}$

$167.56 = \frac{y - 111.92}{x + 2}$

$167.56 \left(x + 2\right) = y - 111.92$

$y = 167.56 x + 167.56 \cdot 2 + 111.92$

$y = 167.56 x + 223 , 21$

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:

${e}^{- 6} = \frac{1}{e} ^ 6$

${e}^{- 6} \cdot {e}^{- 6} = {\left({e}^{- 6}\right)}^{2} = {e}^{- 6 \cdot 2} = {e}^{- 12}$