What is the equation of the line tangent to f(x)=x5−ex at x=2?

1 Answer
Maggie
Mar 6, 2016

y=(80−e2)⋅x+e2−128

exactly or

y=72.61x−120.61

approximately.

Explanation:

You need to take the derivative and the evaluate it at x=2.

We have

f(x)=x5−ex.

We take

df(x)dx=ddx(x5−ex)=ddx(x5)−ddx(ex)

using the Linearity of differentiation.

Now we apply the rules for taking the derivative of powers and exponential functions, Dxxn=nxn−1 and Dxex=ex (isn't Euler's notion so compact?).

df(x)dx=5⋅x4−ex

We need to evaluate this at x=2, because this is the slope of the tangent line at x=2.

m=5⋅24−e2=80−e2≅72.61.

To find the equation of this line we need to know a point on the line, the only point we know is that the line touches the function at
x=2, so (2,(f(2)), is on the line.

We need f(2),

f(2)=25−e2=32−e2≅24.61.

Recall the equation of the line,

y=m⋅x+b,

so we have

32−e2=(80−e2)⋅2+b=160−2⋅e2+b.

Rearranging we get

b=32−e2−160+2e2=e2−128≅−120.61.

Putting everything back into the line we have,

y=(80−e2)⋅x+e2−128

exactly or

y=72.61x−120.61

approximately.

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