To determine which function \( f(x) \) or \( g(x) \) eventually exceeds the other as \( x \) grows larger, we can compare their behaviors as \( x \to \infty \).
The functions given are:
- \( f(x) = 2x^2 + 1 \)
- \( g(x) = 5x + 5 \)
As \( x \) becomes very large, the highest degree term in each function will dominate the behavior of that function.
For \( f(x) \):
- The dominant term is \( 2x^2 \).
For \( g(x) \):
- The dominant term is \( 5x \).
Now let's analyze their growth rates:
- \( f(x) = 2x^2 \) grows quadratically as \( x \to \infty \).
- \( g(x) = 5x \) grows linearly as \( x \to \infty \).
Since quadratic growth (\( x^2 \)) will eventually surpass linear growth (\( x \)) for sufficiently large values of \( x \), we conclude that:
As \( x \) grows larger, \( f(x) \) will eventually exceed \( g(x) \).
To confirm this, we can look for where \( f(x) \) is equal to \( g(x) \):
\[ 2x^2 + 1 = 5x + 5 \]
Rearranging this gives:
\[ 2x^2 - 5x - 4 = 0 \]
Now, we can use the quadratic formula to solve for \( x \):
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
where \( a = 2, b = -5, c = -4 \):
\[ b^2 - 4ac = (-5)^2 - 4(2)(-4) = 25 + 32 = 57 \]
Thus,
\[ x = \frac{5 \pm \sqrt{57}}{4} \]
Calculating these roots gives two values, one of which will be positive. However, what matters here is that for sufficiently large \( x \) (beyond the positive root), \( f(x) \) will indeed exceed \( g(x) \).
In conclusion, \( f(x) \) eventually exceeds \( g(x) \) as \( x \) becomes large.
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