Question

find the equation for the tangent line to f(x)= 3^2x+1 at a=-2

Answer 1

find df/dx

Answer 2

what is that

Answer 3

are you studying calculus ?

Answer 4

yes but this pellet is hard as hell

Answer 5

pellet

Answer 6

can you use the equation editor to put in your equation? I can't tell what is the exponent

Answer 7

is it
\[ f(x)= 3^{2x+1} \]?

Answer 8

\[f(x)= 3^{2x+1} at a=-2\]

Answer 9

when x is in the exponent , think "e"
in other words, we can write
\[ 3= e^{\ln3} \]
and your problem is
\[ f(x) = \left( e^{\ln3}\right)^{2x+1} \\
f(x) = e^{ (2\ln3)x + \ln3}
\]

Answer 10

yea bro i have no clue what this means

Answer 11

You can try Khan's site
http://www.khanacademy.org/math/calculus/differential-calculus/derivative_intro/v/calculus--derivatives-1--new-hd-version

he has lots of stuff on calculus. But if you don't know this, you are not going to learn it by me typing in this box. It will take a lot of work on your part.

Answer 12

just take log both side
logf(x)= log(3)^(2x+1)
logf(x)= (2x+1)*log3=2xlog3+log3
now differentiate with respect to x
on differentiating
1/f(x) * f ' (x)=2*log3
f ' (x)=f(x)*2*log3
but f(x)= 3^2x+1
f ' (x)=3^(2x+1)*2*log3
at x=-2
f ' (-2)=3^(2*(-2)+1)*2*log3=3^(-3)*2*log3=(2log3)/27
this is slope of tangent

Answer 13

f(x)= 3^2x+1
at x=-2
f(-2)=y=3^(-4+1)=1/27
so it is passing through (-2,1/27) and slope=(2log3)/27