1. Given f(x)=3^(x-2), simplify f(x+1)-f(x) into an expression of the form (a)(3^x)
2. Given the function f(x)=e^(x) ln x, write the equation(s) for any and all asymptotes. If the graph has no asymptotes, state that fact.
1. To simplify the expression f(x+1)-f(x), we need to evaluate f(x+1) and f(x) separately, and then subtract the result.
First, let's evaluate f(x+1):
f(x+1) = 3^((x+1)-2) [Substitute x+1 into the given expression for f(x)]
= 3^(x-1)
Now, let's evaluate f(x):
f(x) = 3^(x-2) [Given expression for f(x)]
Substituting both these values back into the expression f(x+1)-f(x), we get:
f(x+1)-f(x) = 3^(x-1) - 3^(x-2)
Now, let's simplify the expression further:
3^(x-1) - 3^(x-2) = (3^x * 3^-1) - (3^x * 3^-2)
= (3^x * 1/3) - (3^x * 1/9)
= 3^x * (1/3 - 1/9) [Factor out the common term, 3^x]
= 3^x * (3/9 - 1/9)
= 3^x * (2/9)
So, f(x+1)-f(x) simplifies to (2/9) * 3^x, which is of the form (a)(3^x), where a = 2/9.
2. To find the asymptotes of the function f(x) = e^x ln x, we need to consider both the exponential function e^x and the natural logarithm function ln x separately.
For the exponential function e^x, there are no asymptotes, as it grows indefinitely in the positive direction.
For the natural logarithm function ln x, there is a vertical asymptote at x = 0. This means that as x approaches 0 from the positive side, ln x approaches negative infinity.
Since f(x) = e^x ln x is a combination of the exponential and logarithmic functions, there is a vertical asymptote at x = 0 due to the logarithmic term.
In summary, the equation for the asymptote of f(x) is x = 0.