Determine if the following functions grow faster, slower, or at the same rate as the function 2e^(2x) as x → ∞.
a. 3.2^(2x)
b. e^(x/6)
c. 1/8 e^(2x)
well, since 3.2 > e, (a) grows faster
e^(x/6) = (e^(1/12))^(2x)
since e^(1/12) < e, (b) grows slower
since 1/8 < 1, (c) grows slower
However, since the ratio is a constant 1/8, you could say that the growth is the same.
To determine if a function grows faster, slower, or at the same rate as another function as x approaches infinity, we can compare the growth rates of their exponents.
In this case, our reference function is 2e^(2x). Let's compare:
a. For the function 3.2^(2x), we can rewrite the base 3.2 as e^(ln(3.2)). Therefore, 3.2^(2x) = (e^(ln(3.2)))^(2x) = e^(2x * ln(3.2)).
Comparing the exponents, we have e^(2x * ln(3.2)) compared to e^(2x). Since ln(3.2) is a positive constant, the exponent of the function 3.2^(2x) grows faster than the exponent of 2e^(2x). Therefore, the function 3.2^(2x) grows faster than 2e^(2x) as x approaches infinity.
b. For the function e^(x/6), the exponent is growing at a slower rate than 2e^(2x) since the denominator is now smaller than 2. Therefore, the function e^(x/6) grows slower than 2e^(2x) as x approaches infinity.
c. For the function 1/8 * e^(2x), the exponent remains the same as 2e^(2x). However, the constant multiplier of 1/8 reduces the growth rate of the function. Therefore, 1/8 * e^(2x) grows slower than 2e^(2x) as x approaches infinity.
In summary:
a. 3.2^(2x) grows faster than 2e^(2x) as x approaches infinity.
b. e^(x/6) grows slower than 2e^(2x) as x approaches infinity.
c. 1/8 e^(2x) grows slower than 2e^(2x) as x approaches infinity.