Question

Both f(x) and g(x) grow as x gets larger and larger. Which function eventually exceeds the other?

f(x) = 2x² + 1

g(x) = 5x + 5

Answer 1

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.