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Find the following for each function: (a) \(f(0)\) (b) \(f(1)\) (c) \(f(-1)\) (d) \(f(-x)\) (e) \(-f(x)\) (f) \(f(x+1)\) (g) \(f(2 x)\) (h) \(f(x+h)\) \(f(x)=\frac{x^{2}-1}{x+4}\)

Short Answer

Expert verified
a) f(0) = -1/4, b) f(1) = 0, c) f(-1) = 0, d) f(-x) = (x^2-1)/(-x+4), e) -f(x) = -(x^2-1)/(x+4), f) f(x+1) = (x(x+2))/(x+5), g) f(2x) = (4x^2-1)/(2x+4), h) f(x+h) = (x^2+2xh+h^2-1)/(x+h+4)

Step by step solution

01

Evaluate for f(0)

To find \( f(0) \), substitute \( x = 0 \) into the function: \[ f(0) = \frac{0^2 - 1}{0 + 4} = \frac{-1}{4} \]
02

Evaluate for f(1)

To find \( f(1) \), substitute \( x = 1 \) into the function: \[ f(1) = \frac{1^2 - 1}{1 + 4} = \frac{0}{5} = 0 \]
03

Evaluate for f(-1)

To find \( f(-1) \), substitute \( x = -1 \) into the function: \[ f(-1) = \frac{(-1)^2 - 1}{-1 + 4} = \frac{1 - 1}{3} = \frac{0}{3} = 0 \]
04

Evaluate for f(-x)

To find \( f(-x) \), substitute \( -x \) for \( x \) in the function: \[ f(-x) = \frac{(-x)^2 - 1}{-x + 4} = \frac{x^2 - 1}{-x + 4} \]
05

Evaluate for -f(x)

To find \( -f(x) \), multiply the function by -1: \[ -f(x) = -\frac{x^2 - 1}{x + 4} \]
06

Evaluate for f(x+1)

To find \( f(x+1) \), substitute \( x+1 \) for \( x \) in the function: \[ f(x+1) = \frac{(x+1)^2 - 1}{x+1 + 4} = \frac{x^2 + 2x + 1 - 1}{x + 5} = \frac{x^2 + 2x}{x + 5} = \frac{x(x+2)}{x+5} \]
07

Evaluate for f(2x)

To find \( f(2x) \), substitute \( 2x \) for \( x \) in the function: \[ f(2x) = \frac{(2x)^2 - 1}{2x + 4} = \frac{4x^2 - 1}{2x + 4} \]
08

Evaluate for f(x+h)

To find \( f(x+h) \), substitute \( x+h \) for \( x \) in the function: \[ f(x+h) = \frac{(x+h)^2 - 1}{x+h + 4} = \frac{x^2 + 2xh + h^2 - 1}{x + h + 4} \]

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Key Concepts

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

Function Evaluation
Function evaluation is about finding the output of a function for specific inputs. The core idea is to replace the variable in the function with a given number or expression.

Let's take the function: \[ f(x) = \frac{x^2 - 1}{x + 4} \] To evaluate the function at a specific point, say at x = 0, you replace \[x \] with \[ 0 \].

So, it looks like this: \[ f(0) = \frac{0^2 - 1}{0 + 4} = \frac{-1}{4} \] This same method is used to find the values at other points, say x = 1 or x = -1.
It's essential to understand that every time you are simply plugging in the value into the function and computing the result.
Function Substitution
Function substitution means replacing the variable in the function with another expression. This is similar to what we did with simple numbers, but now we use expressions.

Let's consider our function: \[ f(x) = \frac{x^2 - 1}{x + 4} \] If we want to evaluate \[ f(x+1) \], we need to substitute \[ x+1 \] for \[ x \] in the function: \[ f(x+1) = \frac{(x+1)^2 - 1}{x+1 + 4} \]

First, expand \[ (x+1)^2 \] to get \[ x^2+2x+1 \]. Then simplify: \[ f(x+1) = \frac{x^2 + 2x + 1 - 1}{x + 5} = \frac{x^2 + 2x}{x + 5} \] By substituting more complex expressions into the function, we can analyze how the function behaves in different situations.
Rational Functions
A rational function is a function that is the ratio of two polynomials. In our example, \[ f(x) = \frac{x^2 - 1}{x + 4} \]

The numerator is \[ x^2 - 1 \], and the denominator is \[ x + 4 \].

Rational functions can have vertical asymptotes, which occur when the denominator equals zero, making the function undefined. In our function, this happens when \[ x + 4 = 0 \] or \[ x = -4 \].

It’s important to note the values that make the denominator zero to understand the function's behavior.
  • Evaluate at different points.
  • Graph the function.
This helps in analyzing and understanding the characteristics of rational functions.
Algebraic Manipulation
Algebraic manipulation involves using algebraic techniques to simplify expressions or solve equations.

In our function example \[ f(x) = \frac{x^2 - 1}{x + 4} \] we often need to simplify expressions. For instance, if we want to find: \[ f(2x) \]
Substitute \[ 2x \] for \[ x \]: \[ f(2x) = \frac{(2x)^2 - 1}{2x + 4} \]
Simplify the numerator \[ (2x)^2 = 4x^2 \], so it becomes \[ f(2x) = \frac{4x^2 -1}{2x + 4} \]

We also often use algebraic techniques to factorize expressions, combine like terms, and cancel out common factors. For example, if you have: \[ -(f(x)) = -\frac{x^2 - 1}{x + 4} \] you are simply multiplying the entire function by -1. Understanding algebraic manipulation is key to correctly evaluating, simplifying, and transforming functions.

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Most popular questions from this chapter

The slope of the secant line containing the two points \((x, f(x))\) and \((x+h, f(x+h))\) on the graph of a function \(y=f(x)\) may be given as \(m_{\mathrm{sec}}=\frac{f(x+h)-f(x)}{(x+h)-x}=\frac{f(x+h)-f(x)}{h} \quad h \neq 0\) (a) Express the slope of the secant line of each function in terms of \(x\) and \(h\). Be sure to simplify your answer. (b) Find \(m_{\text {sec }}\) for \(h=0.5,0.1\), and 0.01 at \(x=1 .\) What value does \(m_{\text {sec }}\) approach as h approaches \(0 ?\) (c) Find an equation for the secant line at \(x=1\) with \(h=0.01\). (d) Use a graphing utility to graph fand the secant line found in part ( \(c\) ) in the same viewing window. Problems \(85-92\) require the following discussion of a secant line. The slope of the secant line containing the two points \((x, f(x))\) and \((x+h, f(x+h))\) on the graph of a function \(y=f(x)\) may be given as \(m_{\mathrm{sec}}=\frac{f(x+h)-f(x)}{(x+h)-x}=\frac{f(x+h)-f(x)}{h} \quad h \neq 0\) (a) Express the slope of the secant line of each function in terms of \(x\) and \(h\). Be sure to simplify your answer. (b) Find \(m_{\text {sec }}\) for \(h=0.5,0.1\), and 0.01 at \(x=1 .\) What value does \(m_{\text {sec }}\) approach as h approaches \(0 ?\) (c) Find an equation for the secant line at \(x=1\) with \(h=0.01\). (d) Use a graphing utility to graph fand the secant line found in part ( \(c\) ) in the same viewing window. \(f(x)=-3 x+2\)

Show that a constant function \(f(x)=b\) has an average rate of change of \(0 .\) Compute the average rate of change of \(y=\sqrt{4-x^{2}}\) on the interval \([-2,2] .\) Explain how this can happen.

Write the standard form of the equation of a circle with center (3,-2) and radius \(r=\frac{\sqrt{6}}{2}\).

Solve: \(|3 x+7|-3=5\)

Explain how the domain of \(g(x)=\sqrt{x}\) compares to the domain of \(g(x-k),\) where \(k \geq 0\).

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