To determine for which pair of functions the exponential function consistently grows at a faster rate than the quadratic function over the interval [tex]\(0 \leq x \leq 5\)[/tex], let's analyze the functions:
1. Quadratic Function: [tex]\( f(x) = x^2 \)[/tex]
2. Exponential Function: [tex]\( g(x) = 2^x \)[/tex]
We need to evaluate these functions at various points within the interval and compare their values. Here's the step-by-step comparison:
### Step-by-Step Comparison:
- At [tex]\( x = 0 \)[/tex]:
- Quadratic: [tex]\( 0^2 = 0 \)[/tex]
- Exponential: [tex]\( 2^0 = 1 \)[/tex]
- Comparison: [tex]\( 1 > 0 \)[/tex]
- At [tex]\( x = 1 \)[/tex]:
- Quadratic: [tex]\( 1^2 = 1 \)[/tex]
- Exponential: [tex]\( 2^1 = 2 \)[/tex]
- Comparison: [tex]\( 2 > 1 \)[/tex]
- At [tex]\( x = 2 \)[/tex]:
- Quadratic: [tex]\( 2^2 = 4 \)[/tex]
- Exponential: [tex]\( 2^2 = 4 \)[/tex]
- Comparison: [tex]\( 4 = 4 \)[/tex]
Here, the exponential function is not strictly greater than the quadratic function.
Since at [tex]\( x = 2 \)[/tex], the two functions are equal, the exponential function does not consistently grow faster than the quadratic function over the entire interval [tex]\(0 \leq x \leq 5\)[/tex].
As a result, the exponential function [tex]\( g(x) = 2^x \)[/tex] does not consistently grow at a faster rate than the quadratic function [tex]\( f(x) = x^2 \)[/tex] over the interval [tex]\(0 \leq x \leq 5\)[/tex]. The specific point of failure is at [tex]\( x = 2 \)[/tex], where [tex]\( g(x) = f(x) = 4 \)[/tex].
