Final answer:
The smallest integer in which the value of f(x) = 2x^(x+1) exceeds the value of g(x) = 2x^2 + x + 1 is x = 3.
Step-by-step explanation:
To find the smallest integer in which the value of f(x) = 2x^(x+1) exceeds the value of g(x) = 2x^2 + x + 1, we can compare the values of the two functions for increasing values of x starting from 1.
Let's evaluate the value of f(x) and g(x) for x = 1:
f(1) = 2(1)^(1+1) = 2*1^2 = 2
g(1) = 2(1)^2 + 1(1) + 1 = 2(1) + 1 + 1 = 2 + 1 + 1 = 4
Since f(1) is smaller than g(1), we need to continue evaluating for larger values of x until we find the smallest integer for which f(x) exceeds g(x).
Evaluating further, f(x) exceeds g(x) for x = 3:
f(3) = 2(3)^(3+1) = 2*3^4 = 2*81 = 162
g(3) = 2(3)^2 + 3(3) + 1 = 2(9) + 9 + 1 = 18 + 9 + 1 = 28
So, the smallest integer for which f(x) exceeds g(x) is x = 3.