help in calculus.....!?

Question

if f (x) = 2^2 and 2^3.03 = 8.168 which of the following is closest to f ' (3)
a. .168
b. .97
c. 1
d. 3
e. 5.6

Answer 1

if f(x) is 2^2, then f(x) is 4 and therefore f'(x) is 0 for all x
you should try rewriting the function

Answer 2

If f(x)=2^2, then your function is a constant (4) and has as derivative 0. I suspect that there are some missing pieces to your question.

Answer 3

I believe you mean:
f(3) = 2^3 or f(x) = 2^x
and
f(3.03) = 2^3.03 = 8.168

You would use the tangent line approximation to find f(3.03), so go in reverse to find the slope "m" or f'(3):

y - y1 = m(x - x1)
y - (2^3) = m(x - 3)

8.168 - 8 = m(3.03 - 3)
m = 0.168 / 0.03
m = 5.6 = f'(3) or E

Answer 4

None of the above!
If f(x ) = 2² = 4
Then f'(x) = 0, so f'(3) = 0.
Did you copy the problem correctly?

Answer 5

if f(x) = 2^2
then f'(x) = 0 for all x

Or is there something you're not telling us?