The function f is given by f(x)=13e−3x+1. What is the equation of the line tangent to the graph of f when x=−ln2?
\boxed{y = -8x - 8 \ln 2 + \frac{11}{3}}
Step 1 :f(x)=13e−3x+1
Step 2 :f′(x)=ddx(13e−3x+1)
Step 3 :f′(x)=−e−3x
Step 4 :x=−ln2
Step 5 :f(−ln2)=13e−3(−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)=−e−3(−ln2)
Step 11 :f′(−ln2)=−e3ln2
Step 12 :f′(−ln2)=−23
Step 13 :f′(−ln2)=−8
Step 14 :y−113=−8(x+ln2)
Step 15 :\boxed{y = -8x - 8 \ln 2 + \frac{11}{3}}