Tangent Line Approximation for f(x) = (4^x)(2^x)

[calculus 1983]

is this correct?

Find the tangent line approximation to f(x) = (4^x)(2^x) for x=2?

Formula: f(x) is approximately equal to f(a) + f'(a)(x - a)

f(x) = (4^x)(2^x)
a = 2

Plug it in:

f(x) = (4^x)(2^x)
f(2) = (4^2)(2^2) = (16)(4) = 64

f'(x) - you are going to use the
Product Rule = (1st)(Derivative of 2nd) + (2nd)(Derivative of 1st)

f'(x) = (4^x)(2^x)(ln2) + (2^x)(4^x)(ln4)
f'(2) = (4^(2))(2^(2))(ln2) + (2^(2))(4^(2))(ln4)
f'(2) = 44.36141956 + 88.72283911

f'(2) = 133.0842587

f'(x) is approximately ( which is 2 of these ~) equal to f(2) + f'(2) x (x-2)

** multiply 133.0842587 by (x - 2)

(4^x)(2^x) is approx. equal to 64 + 133.0842587x - 266.1685173

combine 64 + -266.1685173 = -202.1685173

Therefore, the final answer is:

(4^x)(2^x) is approx. equal to 133.0842587x - 202.1685173

 

[stapel]

Looks good to me!

Note: You can check for "closeness" by graphing the two functions in your calculator (1 < x < 3, -50 < y < 100) to verify that, near x = 2, the two lines are very close.

Eliz.

 

[calculus 1983]

Thank you very much =)
I'm trying to study for an exam and unfortunately do not have any tangent line approximation problems including the product rule, in it. Therefore, I made up that one. Is there any chance you know of any problems i can do similar to this?

-Thanks

 

[morson]

Thank you very much =)
I'm trying to study for an exam and unfortunately do not have any tangent line approximation problems including the product rule, in it. Therefore, I made up that one. Is there any chance you know of any problems i can do similar to this?

-Thanks

Try looking on Google.