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A function \(f\) is given by $$ f(x)=\frac{1}{(x+3)^{2}} $$ This function takes a number \(x,\) adds \(3,\) squares the result, and takes the reciprocal of that result. a) Find \(f(4), f(0), f(a), f(t+4), f(x+h),\) and $$ \frac{f(x+h)-f(x)}{h} $$ b) Note that \(f\) could also be given by $$ f(x)=\frac{1}{x^{2}+6 x+9} $$ Explain what this does to an input number \(x\).

Short Answer

Expert verified
The function calculates values by adding 3 to \(x\), squaring, and taking the reciprocal. For part a: \(f(4)\) is \(\frac{1}{49}\), \(f(0)\) is \(\frac{1}{9}\), expressions for \(f(a), f(t+4), f(x+h)\) and the difference quotient were also derived.

Step by step solution

01

Calculate \(f(4)\)

To find \(f(4)\), substitute \(x = 4\) into the function: \[ f(4) = \frac{1}{(4+3)^{2}} = \frac{1}{7^{2}} = \frac{1}{49}. \]
02

Calculate \(f(0)\)

Substitute \(x = 0\) into the function: \[ f(0) = \frac{1}{(0+3)^{2}} = \frac{1}{3^{2}} = \frac{1}{9}. \]
03

Simplify \(f(a)\)

Substitute \(x = a\) into the function: \[ f(a) = \frac{1}{(a+3)^{2}}. \]
04

Simplify \(f(t+4)\)

Substitute \(x = t+4\) into the function: \[ f(t+4) = \frac{1}{((t+4)+3)^{2}} = \frac{1}{(t+7)^{2}}. \]
05

Simplify \(f(x+h)\)

Substitute \(x = x+h\) into the function: \[ f(x+h) = \frac{1}{((x+h)+3)^{2}} = \frac{1}{(x+h+3)^{2}}. \]
06

Compute \(\frac{f(x+h)-f(x)}{h}\)

We find the difference quotient by calculating \[ \frac{f(x+h) - f(x)}{h} = \frac{\frac{1}{(x+h+3)^{2}} - \frac{1}{(x+3)^{2}}}{h}. \] The common denominator is \((x+h+3)^{2}(x+3)^{2}\), so: \[ \frac{(x+3)^{2} - (x+h+3)^{2}}{h((x+3)^{2}(x+h+3)^{2})}. \] Expand and simplify the numerator: \((x+3)^{2} - (x+3+h)^{2} = -2xh - h^{2} - 6h\). The full expression becomes: \[ \frac{-2xh - h^{2} - 6h}{h((x+3)^{2}(x+h+3)^{2})}, \] which simplifies to \[ \frac{-2x - h - 6}{((x+3)^{2}(x+h+3)^{2})}. \]
07

Explain transformation in \(f(x) = \frac{1}{x^{2} + 6x + 9}\)

This expression expands \((x+3)^{2}\) into \(x^{2} + 6x + 9\), which defines the same function in a polynomial form. It means the function squares the result of adding 3 to \(x\) and then takes the reciprocal, as previously noted. Thus, both expressions represent identical transformations of \(f(x)\).

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Key Concepts

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

Function Analysis
Function analysis involves breaking down a function to understand how changes in the input affect the output. This can include evaluating the function at specific points, understanding the behavior of the function, and interpreting its graphical representation. For the function \( f(x) = \frac{1}{(x+3)^{2}} \), analyzing involves plug-and-play for different values of \( x \) like 4, 0, \( a \), \( t+4 \), or even expressions like \( x+h \):
  • For \( x = 4 \), it simplifies to \( \frac{1}{49} \).
  • For \( x = 0 \), it results in \( \frac{1}{9} \).
  • At \( x = a \) or when \( x = t+4 \), the function gives a symbolic expression which helps in analyzing the overall behavior.
By evaluating these different cases, we can infer the function's characteristics, like its domain, range, and any symmetry or asymptotic behavior due to the squared denominator.
Difference Quotient
The difference quotient is a formula that gives the average rate of change of the function over a particular interval. It's mainly represented as \( \frac{f(x+h) - f(x)}{h} \). Calculating this for our function \( f(x) = \frac{1}{(x+3)^{2}} \) involves a bit of algebra:1. Substitute \( x+h \) and \( x \) into the function resulting in \( f(x+h) = \frac{1}{(x+h+3)^{2}} \) and \( f(x) = \frac{1}{(x+3)^{2}} \).2. Compute \( \frac{1}{(x+h+3)^{2}} - \frac{1}{(x+3)^{2}} \), which requires finding a common denominator before simplifying.3. Simplify the expression which eventually reveals the "growth" or "shrink" rate of the function for small increments of \( h \).The difference quotient is the backbone of deriving the derivative, thus highlighting how crucial it is for understanding calculus.
Polynomial Transformation
Understanding polynomial transformation helps in recognizing how functions can look different algebraically but behave the same. The function \( f(x) = \frac{1}{(x+3)^{2}} \) and \( f(x) = \frac{1}{x^{2} + 6x + 9} \) involve a transformation that utilizes expanding and simplifying polynomials.- \( (x+3)^{2} \) expands to \( x^{2}+6x+9 \), showcasing its polynomial form.- This transformation is crucial because it may simplify or clarify the appearance of complex expressions, like making addition or multiplication of polynomials more straightforward.- The operation doesn't change the function's behavior or its intrinsic properties but provides different perspectives for solving problems.Such transformations are vital in simplifying algebraic expressions and solving calculus problems efficiently.
Rational Functions
Rational functions are functions that are represented as the ratio of two polynomials. For example, \( f(x) = \frac{1}{(x+3)^{2}} \) is a rational function with a constant numerator and a squared polynomial denominator.- These functions often have vertical asymptotes where the denominator equals zero.- The graph of \( f(x) \) would display a key feature: it approaches infinity as \( x \) gets closer to a point where \( (x+3) \) equals zero, indicating a vertical asymptote at \( x = -3 \).- Additionally, rational functions can exhibit holes if there are common factors in the numerator and denominator, although our current function does not display this.Rational functions are a crucial concept in calculus for studying asymptotic behavior, especially in limits and graphing contexts.

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Most popular questions from this chapter

The annual interest rate \(r,\) when compounded more than once a year, results in a slightly higher yearly interest rate; this is called the annual (or effective) yield and denoted as Y. For example, \$1000 deposited at 5\%, compounded monthly for 1 yr \((12\) months \(),\) will have a value of \(A=1000\left(1+\frac{0.05}{12}\right)^{12}=\$ 1051.16 .\) The interest earned is \(\$ 51.16 / \$ 1000,\) or \(0.05116,\) which is \(5.116 \%\) of the original deposit. Thus, we say this account has a yield of \(Y=0.05116,\) or \(5.116 \% .\) The formula for annual yield depends on the annual interest rate \(r\) and the compounding frequency \(n:\) \(Y=\left(1+\frac{r}{n}\right)^{n}-1.\) For Exercises 41-48, find the annual yield as a percentage, to two decimal places, given the annual interest rate and the compounding frequency. Annual interest rate of \(3.75 \%,\) compounded weekly

Refer to Example \(14 .\) The home range, in hectares, of an omnivorous mammal of mass \(w\) grams is given by $$H(w)=0.059 w^{0.92}$$ (Source: Harestad, A. S., and Bunnel, F. L., "Home Range and Body Weight-A Reevaluation," Ecology, Vol. 60, No. 2 (April, 1979), pp. 405-418.) Complete the table of approximate function values and graph the function. $$ \begin{array}{|c|c|c|c|c|c|c|c|c|} \hline w & 0 & 1000 & 2000 & 3000 & 4000 & 5000 & 6000 & 7000 \\ \hline H(w) & 0 & 34.0 & & & & & & \\ \hline \end{array} $$

A function \(f\) takes a number \(x\), multiplies it by \(3,\) and then adds \(6,\) while a function \(g\) takes a number \(x\), adds \(a\) to it, and then multiplies the result by \(3 .\) Find \(a\) if \(f\) and \(g\) are the same function.

Determine the domain of each function. $$ f(x)=\frac{x^{2}-4}{x+2} $$

Rewrite each of the following as an equivalent expression with rational exponents. $$ \sqrt[4]{b^{2}}, \quad b \geq 0 $$

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