/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 17 For the functions \(f(x)=3^{x}, ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

For the functions \(f(x)=3^{x}, g(x)=\left(\frac{1}{16}\right)^{x},\) and \(h(x)=10^{x+1},\) find the function value at the indicated points. $$g\left(-\frac{1}{2}\right)$$

Short Answer

Expert verified
The function value is 4.

Step by step solution

01

Identify the function

We are given the function to evaluate: \( g(x) = \left(\frac{1}{16}\right)^x \). This is an exponential function with base \( \frac{1}{16} \).
02

Substitute the value into the function

Let's substitute \( x = -\frac{1}{2} \) into the function: \[ g\left(-\frac{1}{2}\right) = \left( \frac{1}{16} \right)^{-\frac{1}{2}} \].
03

Rewrite the expression using positive exponents

We can convert the negative exponent by taking the reciprocal of the base: \[ \left( \frac{1}{16} \right)^{-\frac{1}{2}} = \left( 16 \right)^{\frac{1}{2}} \].
04

Evaluate the expression

The expression \( 16^{\frac{1}{2}} \) means the square root of 16. Since \( \sqrt{16} = 4 \), we find that the value is \( 4 \).
05

Conclusion

The function value of \( g \left( -\frac{1}{2} \right) \) is \( 4 \).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Function Evaluation
Function evaluation is a fundamental concept in mathematics. When we evaluate a function, we substitute a specific value in place of the variable. In the context of exponential functions, like the given function \( g(x) = \left(\frac{1}{16}\right)^x \), the process involves substituting the given \( x \) value into the function and simplifying.
  • Identify the function and the evaluation point.
  • Substitute the \( x \) value into the function.
  • Simplify the expression to get the result.For our specific example, we substitute \(-\frac{1}{2}\) into the function \( g(x) \). This results in \( g\left(-\frac{1}{2}\right) = \left( \frac{1}{16} \right)^{-\frac{1}{2}} \).
The resulting expression can then be simplified by employing the concepts of negative exponents and square roots.
Negative Exponents
Negative exponents can seem tricky at first, but they follow a simple rule: a negative exponent indicates the reciprocal of the base raised to the opposite positive exponent. For example, a function with a negative exponent like \( \left(\frac{1}{16}\right)^{-\frac{1}{2}} \) is equivalent to \( \left(16\right)^{\frac{1}{2}} \).

Steps to Simplify Negative Exponents:

  • Identify the base and the negative exponent.
  • Take the reciprocal of the base. The reciprocal of a number \( a \) is \( \frac{1}{a} \).
  • Change the sign of the exponent to positive.
In this case, the reciprocal of \( \frac{1}{16} \) is 16, which gives us \( 16^{\frac{1}{2}} \) as the equivalent expression with a positive exponent.
Square Roots
Square roots are a special mathematical operation that find a number which, when multiplied by itself, equals the original number. When looking at a function like \( 16^{\frac{1}{2}} \), it is expressing the square root: \( \sqrt{16} \).

Understanding the Square Root:

  • The square root of a number \( a \) is denoted as \( \sqrt{a} \).
  • If \( a = b^2 \), then \( \sqrt{a} = b \).
  • For our example, \( \sqrt{16} = 4 \) because \( 4 \times 4 = 16 \).
By understanding and applying the concept of square roots, we completed the evaluation of our function, yielding the final value of 4.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Study anywhere. Anytime. Across all devices.