Chapter 3: Problem 11
Exer. 11-14: If \(a\) is a positive real number, find (a) \(g\left(\frac{1}{a}\right)\) (b) \(\frac{1}{g(a)}\) (c) \(g(\sqrt{a})\) (d) \(\sqrt{g(a)}\) $$ g(x)=4 x^{2} $$
Short Answer
Expert verified
(a) \(\frac{4}{a^2}\), (b) \(\frac{1}{4a^2}\), (c) \(4a\), (d) \(2a\).
Step by step solution
01
Evaluate \(g\left(\frac{1}{a}\right)\)
Given the function \(g(x) = 4x^2\), to find \(g\left(\frac{1}{a}\right)\), substitute \(\frac{1}{a}\) for \(x\) in the function:\[g\left(\frac{1}{a}\right) = 4\left(\frac{1}{a}\right)^2 = 4 \cdot \frac{1}{a^2} = \frac{4}{a^2}.\]
02
Evaluate \(\frac{1}{g(a)}\)
The function \(g(x) = 4x^2\) means that \(g(a) = 4a^2\). Therefore, to find \(\frac{1}{g(a)}\), compute: \[\frac{1}{g(a)} = \frac{1}{4a^2}.\]
03
Evaluate \(g(\sqrt{a})\)
For \(g(x) = 4x^2\), substitute \(\sqrt{a}\) for \(x\): \[g(\sqrt{a}) = 4(\sqrt{a})^2 = 4a.\] The square root and the square cancel out, leaving you with \(4a\).
04
Evaluate \(\sqrt{g(a)}\)
First find \(g(a)\) which is \(4a^2\). Then compute the square root of that value: \[\sqrt{g(a)} = \sqrt{4a^2}.\] This simplifies to \[2a.\]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Quadratic Functions
A quadratic function is a type of polynomial function that takes the form \( f(x) = ax^2 + bx + c \). Here, \( a \), \( b \), and \( c \) are constants, and \( a eq 0 \). In the exercise, the quadratic function is given as \( g(x) = 4x^2 \). This is a simple quadratic function where:
- The coefficient \( a \) is 4.
- There is no linear term (\( bx \)) or constant term (\( c \)).
Substitution
Substitution is a method used to evaluate functions by replacing variables with specific values. In our exercise, we substitute values into the function \( g(x) = 4x^2 \). The key is to carefully replace \( x \) with the given expression and perform the operations step by step. For instance:
- To find \( g\left(\frac{1}{a}\right) \), replace \( x \) with \( \frac{1}{a} \) to get \( 4\left(\frac{1}{a}\right)^2 = \frac{4}{a^2} \).
- For \( g(\sqrt{a}) \), substitute \( x \) with \( \sqrt{a} \), resulting in \( 4(\sqrt{a})^2 = 4a \).
Square Roots
The square root is a fundamental mathematical operation that finds a number whose square equals the given number. For example, the square root of 4 is 2 because \( 2^2 = 4 \). In our exercise, we see square roots used several times:
- When evaluating \( g(\sqrt{a}) \), where the square root of \( a \) is squared, simplifying to \( a \).
- When finding \( \sqrt{g(a)} \), which simplifies to \( \sqrt{4a^2} = 2a \).
Inverse Functions
Inverse functions reverse the effect of the original function. If you have a function \( f(x) \), its inverse \( f^{-1}(x) \) will map the output back to the original input. However, only functions that are one-to-one (bijective) have inverses. In the context of the function \( g(x) = 4x^2 \), consider its inverse. The function \( g(x) \) does not have a straightforward inverse over all real numbers due to the squaring operation, which is not one-to-one for all real numbers, but it could be considered on restricted domains.For example, given \( g(x) = 4x^2 \), the inverse could be determined by solving \( y = 4x^2 \) for \( x \). After which, we can conclude an inverse involves a square root, but it is only valid for positive values of \( x \), reflecting the restriction needed for a true inverse. This demonstrates how useful but intricate inverses can be, requiring a deep understanding of both the function and its domain.