Problem

The function f is given by f(x)=13e3x+1. What is the equation of the line tangent to the graph of f when x=ln2?

Answer

Expert–verified
Hide Steps
Answer

\boxed{y = -8x - 8 \ln 2 + \frac{11}{3}}

Steps

Step 1 :f(x)=13e3x+1

Step 2 :f(x)=ddx(13e3x+1)

Step 3 :f(x)=e3x

Step 4 :x=ln2

Step 5 :f(ln2)=13e3(ln2)+1

Step 6 :f(ln2)=13e3ln2+1

Step 7 :f(ln2)=13(23)+1

Step 8 :f(ln2)=83+1

Step 9 :f(ln2)=113

Step 10 :f(ln2)=e3(ln2)

Step 11 :f(ln2)=e3ln2

Step 12 :f(ln2)=23

Step 13 :f(ln2)=8

Step 14 :y113=8(x+ln2)

Step 15 :\boxed{y = -8x - 8 \ln 2 + \frac{11}{3}}

link_gpt