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#.