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

1 Answer
Jan 12, 2018

Obtain the slope by taking the derivative at x = -2, then solve for f(-2), to later solve for b, and simplifying the equation after using the slope-intercept form, giving a final result of y = 168x + 71e^6 x + 224 + 124e^6.

Explanation:

So, we have

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

And want to find the equation of the tangent line at x = -2. We do know that the slope of this tangent line would be the derivative of the function:

(df)/(dx) = d/dx (14x^3 - (4x^2 - x)/(e^(3x)))

Using the sum rule, we could take the derivative of each term (and take the subtraction out):

(df)/(dx) = d/dx (14x^3) - d/dx((4x^2 - x)/(e^(3x)))

First taking the derivative of the term on the left:

d/dx (14x^3) = 14 d/dx(x^3) = 14 * 3x^2 = 42x^2

Putting that back in:

(df)/(dx) = 42x^2 - d/dx((4x^2 - x)/(e^(3x)))

As for this second function... let's set a function g(x) where

g(x) = (4x^2 - x)/(e^(3x))

So that

(df)/(dx) = 42x^2 - (dg)/(dx)

And we have to take the derivative of g(x). We will first decompose it (break it down) into two smaller functions, one of the numerator, p(x), and the other of the denominator, q(x):

p(x) = 4x^2 - x

q(x) = 1/(e^(3x))

So that

g(x) = p(x)q(x)

And, taking the derivative, by the product rule, we have

dg = p(x)dq + q(x)dp

"Dividing" the differentials by dx:

(dg)/(dx) = p(x)(dq)/(dx) + q(x)(dp)/(dx)

Substituting what we already have:

(dg)/(dx) = (4x^2 - x)(dq)/(dx) + (1/(e^(3x)))(dp)/(dx)

Now, we need to take the derivatives of p(x) and q(x). Starting from p(x), we can separate its two terms using the sum rule, and take out the subtraction:

(dp)/(dx) = d/dx (4x^2) - d/dx (x)

Solve for each term:

(dp)/(dx) = 4 d/dx (x^2) - d/dx (x)

(dp)/(dx) = 4 * 2x - 1

(dp)/(dx) = 8x - 1

Now put it back:

(dg)/(dx) = (4x^2 - x)(dq)/(dx) + (1/(e^(3x)))(8x - 1)

Let's now solve for the derivative of q(x):

(dq)/(dx) = d/dx (1/(e^(3x)))

We can use the chain rule here. Decompose q(x) into several parts:

q_1(x) = 3x

q_2(x) = e^x

q_3(x) = 1/x

So that q(x) = q_3(q_2(q_1(x))). Let's first take the derivative of q_1(x):

(dq_1)/(dx) = d/dx (3x) = 3

Also solve for the differential by "multiplying" by dx:

(dq_1)/(dx) = 3 rarr dq_1 = 3 dx

Now, take the derivative of q_2(x) with respect to q_1(x), which means to shove q_1 as the input:

(dq_2)/(dq_1) = d/(dq_1) (e^(q_1)) = e^(q_1)

Solve for the differential dq_2:

(dq_2)/(dq_1) = e^(q_1) rarr dq_2 = e^(q_1) dq_1

And evaluate q_1, as well as dq_1 as we have solved for earlier:

dq_2 = e^(3x) * 3dx = 3e^(3x)dx

Finally, "divide" by dx to solve for the derivative:

(dq_2)/(dx) = 3e^(3x)

Next, the derivative of q_3(x) with respect to q_2(x):

(dq_3)/(dq_2) = d/(dq_2) (1/x) = - 1/((q_2)^2)

Solving for the differential:

(dq_3)/(dq_2) = - 1/((q_2)^2) rarr dq_3 = - 1/((q_2)^2) dq_2

And evaluating:

(dq_3)/(dq_2) = - 1/((q_2)^2) rarr dq_3 = - 1/((e^(q_1))^2) * 3e^(3x) dx

dq_3 = - (3e^(3x))/((e^(q_1))^2) dx

Again, for q_1:

dq_3 = - (3e^(3x))/((e^(3x))^2) dx

Simplifying further:

dq_3 = - (3)/(e^(3x)) dx

Finally dividing through:

(dq_3)/(dx) = -(3)/(e^(3x))

Since we differentiated all the way, this is the derivative of q(x) as a whole:

(dq)/(dx) = -(3)/(e^(3x))

Now let's put it back to the derivative of g(x):

(dg)/(dx) = (4x^2 - x)(dq)/(dx) + (1/(e^(3x)))(8x - 1)

(dg)/(dx) = (4x^2 - x)(-(3)/(e^(3x))) + (1/(e^(3x)))(8x - 1)

And simplify:

(dg)/(dx) = (4x^2 - x)(-(3)/(e^(3x))) + (8x - 1)/(e^(3x))

(dg)/(dx) = -(3(4x^2 - x))/(e^(3x)) + (8x - 1)/(e^(3x))

(dg)/(dx) = (3(-4x^2 + x))/(e^(3x)) + (8x - 1)/(e^(3x))

(dg)/(dx) = (-12x^2 + 3x)/(e^(3x)) + (8x - 1)/(e^(3x))

(dg)/(dx) = ((-12x^2 + 3x) + (8x - 1))/(e^(3x))

(dg)/(dx) = (-12x^2 + 11x - 1)/(e^(3x))

Now that we have the derivative of g(x), let's put it back into the derivative of f(x):

(df)/(dx) = 42x^2 - (dg)/(dx)

(df)/(dx) = 42x^2 - (-12x^2 + 11x - 1)/(e^(3x))

(df)/(dx) = 42x^2 + (12x^2 - 11x + 1)/(e^(3x))

The derivative is the slope of the tangent line for any x. Since we have x = -2, substitute:

rarr 42(-2)^2 + (12(-2)^2 - 11(-2) + 1)/(e^(3(-2)))

= 42(4) + (12(4) + 22 + 1)/(e^(-6))

= 168 + (48 + 23)(e^6)

= 168 + 71e^6

= 168 + 71e^6

And that, my friends, is the slope of our tangent line. Hmmh, but if we were to use the slope-intercept form y = mx + b, and we know m = 168 + 71e^6 (for x = -2), what is b?

Well, we could take a point on the graph (yes, our original function's graph, since this tangent line should touch it) and put in its x and y coordinates. We could use x = -2, but we'd need to solve for f(x) to get y:

f(-2) = 14(-2)^3 - (4(-2)^2 - (-2))/(e^(3(-2)))

f(-2) = 14(-8) - (4(4) + 2)/(e^(-6))

f(-2) = -112 - (16 + 2)(e^6)

f(-2) = -112 - 18e^6

Alright, so y = -112 - 18e^6. Let's substitute for what we have:

y = mx + b rarr -112 - 18e^6 = (168 + 71e^6)(-2) + b

And solve for b:

b = -112 - 18e^6 + (168 + 71e^6)(2)

b = -112 - 18e^6 + 336 + 142e^6

b = 224 + 124e^6

Now that we know m and b, we could put them back in y = mx + b to get:

y = (168 + 71e^6)x + 224 + 124e^6

y = 168x + 71e^6 x + 224 + 124e^6

Which is the equation of our tangent line at x = -2.