Question

Can someone walk me through how to do this

f(x)=2^x
and
f(x)=2^x+1

Answer 1

I'm not sure on where to start

Answer 2

What are you trying to do with the two functions and is the \(x+1\) in the exponent?

Answer 3

Graph and yes it is

Answer 4

Here are the things to consider and guidelines when you have to sketch any function:

A. Domain. What are the domains of the function in interest?
B. Symmetry. Is it even or odd or periodic (ie sinusoidal graph) or none.
C. Asymptotes and bound. Are there asymptotes? Is it bounded? If so, where is it bounded?
D. Transformations - shifts, stretches, and compression. (this is useful in your example). The second function is f(x+1) for the first function. This means that the second function is shifted one point to the left.
E. Intercepts - find the x and y intercepts
F. Critical points - find the critical points by setting the first derivative of the function to 0.
G. Local Extrema - this is in continuation of C.P. Plug in the CP into the second derivative. f'' < 0 means maxima and f''x > 0 means minima. If it is 0, then it is not a local extrema and just a critical point.
H. Concavity - Let I be a subset of the domain of f. If at I the second derivative is less that 0, the I area is concave down and visa versa.
I. Inflection - Set f '' x to 0 and find the possible x value. This is the point of inflection.

Answer 5

So in your case, closely examine the domain, axis intercepts, and transformations

Answer 6

Or also choose test points to start you off: -10, -5, -1, 0, 1, 5, 10

Answer 7

So the domains would be like all the real numbers right?

Answer 8

Yes.

Answer 9

Also think of the range. That would account for whether the function is bounded or not

Answer 10

Do you get what you need to do? I'll give an example

Say we have \( f(x) = 3^{x-1}\)
Go in order.
A. Domain. Well the the limit of x approaching negative infinity is 0, so negative numbers are in the domain. 0 is also in the domain. The limit of x approaching positive infinity is simply infinity, so in essence, the domain is all real numbers.

B. Symmetry. Well it would not be any if we plug in f(-x). And it is not a periodic function

C. Asymptotes. None. But there is a lower bound. The function has no negative outputs.

D. Transformations. I can notice a composite function and express f(x) as the composite h(g(x)) for two different functions: \( h(x) = 3^x~and~g(x) = x-1\). If we take h(g(x)) we get f(x). Thus, we can say \(f(x) = h(x-1)\). In otherwords, h(x) was shifted 1 unit to the right.

E. Intercepts:
For the y-intercept, I set x as 0 and obtain y as 1/3.
For the x-intercept, there is none as the function never actually hits the x-axis, but it only gets fractionally closer. This is why the limit of f(x) is 0 even though there is no true intersection.

We can skip the remainder of the steps because the derivative of the function would not cross the x-axis either.

Answer 11

Ok I understand

Let's say we have this
f(x)=5^x
and I graphed it and the points were

(-2,1/25)
(-1,1/5)
(0,1)
1,5)
(2,25) and on..





(couldn't graph all the points)

Would this be correct?

Answer 12

Yes it's right. You can check your graph on desmos.com