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Chapter 2: Problem 3

Find the indicated function values. If \(f(x)=\sqrt{x+1} ; \quad f(-1), f(0), f(3),\) and \(f(5)\)

Short Answer

Expert verified
\( f(-1) = 0, f(0) = 1, f(3) = 2, f(5) \approx 2.449 \)

Step by step solution

01

Understanding the Function

The given function is a square root function expressed as \( f(x) = \sqrt{x + 1} \). This function takes an input \( x \) and adds \( 1 \) to it before applying the square root.
02

Calculating \( f(-1) \)

To find \( f(-1) \):Substitute \( x = -1 \) into the function:\[ f(-1) = \sqrt{-1 + 1} = \sqrt{0} = 0 \]
03

Calculating \( f(0) \)

To find \( f(0) \):Substitute \( x = 0 \) into the function:\[ f(0) = \sqrt{0 + 1} = \sqrt{1} = 1 \]
04

Calculating \( f(3) \)

To find \( f(3) \):Substitute \( x = 3 \) into the function:\[ f(3) = \sqrt{3 + 1} = \sqrt{4} = 2 \]
05

Calculating \( f(5) \)

To find \( f(5) \):Substitute \( x = 5 \) into the function:\[ f(5) = \sqrt{5 + 1} = \sqrt{6} \approx 2.449 \]

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Key Concepts

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

Square Root Function
The square root function is a mathematical operation that is often denoted by the symbol \( \sqrt{} \). In this specific case, we're dealing with the function \( f(x) = \sqrt{x + 1} \). This function is known as a "square root function" because it involves taking the square root of an expression.
  • For any positive number inside the square root, it finds the number that, when multiplied by itself, equals the original number.
  • For the square root function, if the inside value is zero, the result is zero, just like \( \sqrt{0} = 0 \).
  • Negative values inside a square root are not defined in the real number system without the use of imaginary numbers. However, in this function \( x + 1 \) must be zero or greater to ensure the values are real.
Understanding how this works is key to evaluating a function that contains a square root.
Function Values
Evaluating function values involves determining the output of a function for particular inputs. For the given function \( f(x) = \sqrt{x + 1} \), we are asked to find specific outputs.
  • \( f(-1) \): Substitute \( x = -1 \) into the function and calculate. As shown, \( f(-1) = \sqrt{-1 + 1} = \sqrt{0} = 0 \).
  • \( f(0) \): Substitute \( x = 0 \), therefore \( f(0) = \sqrt{0 + 1} = \sqrt{1} = 1 \).
  • \( f(3) \): Here, \( x = 3 \), so \( f(3) = \sqrt{3 + 1} = \sqrt{4} = 2 \).
  • \( f(5) \): Finally, substitute \( x = 5 \) to get \( f(5) = \sqrt{5 + 1} = \sqrt{6} \approx 2.449 \).
Each of these values shows how different inputs affect the function, prompting different square roots as outputs.
Substitution Method
The substitution method is a simple technique used to evaluate functions at specific points. This method involves replacing the variable, often \( x \), with a certain number to find a function's value.
  • Start by identifying the function, such as \( f(x) = \sqrt{x + 1} \).
  • Choose an input value, like \( x = -1 \), \( 0 \), \( 3 \), or \( 5 \), depending on what's asked.
  • Replace \( x \) with these numbers, one at a time, in the function.
  • Simplify the resulting expression to find the value of the function at that input.
This method helps systematically solve for specific points on the function graph and understand how inputs translate into outputs.

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Most popular questions from this chapter

Consider the semicircle whose equation is \(f(x)=\sqrt{1-x^{2}} .\) Discuss: How can the derivative \(f(x)\) be found using only the geometric fact that the radius of a circle is perpendicular to the tangent line at a point \((x, y)\) on the circle?

The given function \(f\) is one-to one. Find \(f^{-1}\). Sketch the graphs of \(f\) and \(f^{-1}\) on the same rectangular coordinate system. $$ f(x)=1-x^{3} $$

The given function \(f\) is one-toone. The domain and range of \(f\) is given. Find \(f^{-1}\) and give its domain and range. $$ f(x)=\frac{2}{\sqrt{x}}, \quad x>0, y>0 $$

Use the appropriate derivatives obtained in Problems 11-26. For the given function, find the point of tangency and slope of the tangent line at the indicated value of \(x .\) Find an equation of the tangent line at that point. $$ f(x)=x+\frac{1}{x}, \quad x=\frac{1}{2} $$

In Problems 47 and 48 , find the value of the constant \(k\) such that the given piecewise-defined function \(f\) is continuous at \(x=2\). That is, the graph of \(f\) has no holes, gaps, or breaks in its graph at \(x=2\). $$ f(x)=\left\\{\begin{array}{ll} \frac{1}{2} x+1, & x \leq 2 \\ k x, & x>2 \end{array}\right. $$
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