Problem 23
Question
Comparing Growth Which function becomes larger for \(0 \leq x \leq 10: f(x)=2^{x}\) or \(g(x)=x^{2} ?\)
Step-by-Step Solution
Verified Answer
For \( 0 \leq x \leq 10 \), \( f(x) = 2^x \) becomes larger than \( g(x) = x^2 \) starting from \( x = 5. \)
1Step 1: Understanding the Functions
We need to compare the growth of two functions: \( f(x) = 2^x \) and \( g(x) = x^2 \) for values of \( x \) ranging from 0 to 10. \( f(x) = 2^x \) is an exponential function while \( g(x) = x^2 \) is a quadratic function.
2Step 2: Evaluate Both Functions at Critical Points
Evaluate both functions at critical points, such as \( x = 0, x = 1, x = 2, \) and continue towards \( x = 10 \):- At \( x = 0 \), \( f(x) = 2^0 = 1 \) and \( g(x) = 0^2 = 0 \).- At \( x = 1 \), \( f(x) = 2^1 = 2 \) and \( g(x) = 1^2 = 1 \).- At \( x = 2 \), \( f(x) = 2^2 = 4 \) and \( g(x) = 2^2 = 4 \).
3Step 3: Compare Both Functions for Increasing x
Continue evaluating for higher \( x \):- At \( x = 3 \), \( f(x) = 2^3 = 8 \) while \( g(x) = 3^2 = 9 \). Here, \( g(x) > f(x) \).- At \( x = 4 \), \( f(x) = 2^4 = 16 \) and \( g(x) = 4^2 = 16 \). Both are equal.- At \( x = 5 \), \( f(x) = 2^5 = 32 \) and \( g(x) = 5^2 = 25 \). Now, \( f(x) > g(x) \).- Continue similarly for \( x = 6, 7, 8, 9, 10 \), noting that \( f(x) \) continues to grow faster.
4Step 4: Conclusion on Growth within Range
Around \( x = 5 \), the exponential function \( f(x) = 2^x \) overtakes the quadratic function \( g(x) = x^2 \) and grows faster for all points \( x > 4 \) up to 10.
Key Concepts
Quadratic FunctionsFunction ComparisonFunction Growth
Quadratic Functions
Quadratic functions are an essential concept in mathematics and appear in various real-world situations. The general form of a quadratic function is \( g(x) = ax^2 + bx + c \), where \( a \), \( b \), and \( c \) are constants, and \( a eq 0 \).The quadratic function \( g(x) = x^2 \) given in the exercise is a simple form, with \( a = 1 \) and \( b = c = 0 \).
- The graph of a quadratic function is a parabola, which can open upwards or downwards depending on the sign of \( a \). In this case, it opens upwards.
- The vertex of a parabola \( f(x) = x^2 \) is at the origin \((0, 0)\).
- Quadratic functions are symmetrical about a vertical line through their vertex.
Function Comparison
Comparing functions involves analyzing their behavior over a specific range of values.In the provided exercise, we compare the quadratic function \( g(x) = x^2 \) with the exponential function \( f(x) = 2^x \).
- Initially, at \( x = 0 \), the exponential function has a higher value, \( f(x) = 1 \) while \( g(x) = 0 \).
- At \( x = 2 \), both functions equate, \( f(x) = g(x) = 4 \).
- For \( x = 3 \) and \( x = 4 \), the quadratic function takes the lead.
Function Growth
Understanding the growth of functions helps us predict their long-term behavior. Quadratic functions, like \( g(x) = x^2 \), grow at a polynomial rate, characterized by how their outputs become larger as \( x \) increases. Exponential functions, such as \( f(x) = 2^x \), grow very quickly as \( x \) becomes larger. They start slower than quadratic functions initially but eventually overshadow them due to their rapid rate of increase:
- Exponential growth implies multiplicative changes. Each increase by 1 in \( x \) results in the function's value doubling.
- While quadratic functions add a linear sequence of growth (like \( x, 2x, 3x, ... \)), exponential growth stacks that on top, creating a much steeper curve.
Other exercises in this chapter
Problem 22
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Simplify the expression. $$ 7^{\log _{7} 2 x} $$
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Determine if \(f\) is one-to-one. You may want to graph \(y=f(x)\) and apply the horizontal line test. $$ f(x)=2 x-7 $$
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Exercises \(11-30:\) Use \(f(x)\) and \(g(x)\) to find a formula for each expression. Identify its domain. (a) \((f+g)(x)\) (b) \((f-g)(x)\) (c) \((f g)(x)\) (d
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