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Prove that if \(a^{2}+b c \neq 0\), then the graph of $$ f(x)=\frac{a x+b}{c x-a} $$ is symmetric about the line \(y=x\)

Short Answer

Expert verified
The function is symmetric about the line \( y = x \).

Step by step solution

01

Find the Inverse Function

To see if the function is symmetric with respect to the line \( y = x \), we need to determine its inverse. Start by setting \( y = \frac{ax+b}{cx-a} \). Switch \( x \) and \( y \) to find the inverse: \( x = \frac{ay+b}{cy-a} \). Solve this equation for \( y \).
02

Solve for y

To express \( y \) in terms of \( x \), first multiply both sides by \( cy-a \): \( x(cy-a) = ay+b \). This expands to: \( xcy - xa = ay + b \). Gather terms involving \( y \): \( xcy - ay = b + xa \). Factor \( y \) out: \( y(xc-a) = b+xa \). So, \( y = \frac{b+xa}{xc-a} \).
03

Verify Symmetry

For symmetry about the line \( y=x \), it must be the case that \( f^{-1}(x) = f(x) \). We found that \( f^{-1}(x) = \frac{b+xa}{xc-a} \). Check to see if this equals the original function \( f(x) = \frac{ax+b}{cx-a} \). Notice that if we substitute \( x \) with \( y \) in \( f(x) \), we have \( f(y) = \frac{ay+b}{cy-a} = \frac{b+xa}{xc-a} \). By substitution, it's shown that \( f^{-1}(x) = f(x) \).
04

Conclude Symmetry

Since the expression \( f(y) = f(x) \) holds true upon proper substitutions, and considering the condition \( a^2 + bc eq 0 \) prevents indeterminate forms, the function \( f(x) = \frac{ax+b}{cx-a} \) is shown to be symmetric about the line \( y = x \).

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Key Concepts

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

Symmetry in Graphs
The symmetry of a graph refers to its mirror-like property. In this context, the function we are examining is symmetrical along the line \( y = x \). This means that if you flip the graph over this line, it would look exactly the same.

To determine such symmetry, one powerful method is to check if the function is its own inverse. Remember, a function \( f(x) \) is symmetric with respect to the line \( y = x \) if \( f^{-1}(x) = f(x) \).
  • First, we find the inverse by swapping \( x \) and \( y \) in the function and solving for \( y \).
  • If the resulting expression is the same as the original function, symmetry about \( y = x \) is confirmed.
This was shown in our original problem when the inverse function matched the original function when we substituted back, confirming the symmetry about the line \( y = x \).
Rational Functions
Rational functions consist of ratios of polynomials. They take the form \( \frac{P(x)}{Q(x)} \), where \( P(x) \) and \( Q(x) \) are polynomials. Understanding these functions is vital because they often appear in mathematical modeling and various equations in science and engineering.

The function \( f(x) = \frac{ax+b}{cx-a} \) is a classic example of a rational function. Here:
  • The numerator is \( ax + b \), a linear polynomial.
  • The denominator is \( cx - a \), another linear polynomial.
Rational functions can exhibit interesting characteristics such as asymptotes, where the function behaves very differently around certain points. The denominator must not be zero, since division by zero is undefined.

In the given problem, the condition \( a^2 + bc eq 0 \) helps ensure that the function’s form remains valid, avoiding undefined expressions and maintaining the integrity of the graph and its properties.
Function Transformation
Function transformation involves changing the position, shape, or orientation of the graph of a function. These changes or "transformations" can include shifts, stretches, reflections, and more.

With rational functions, transformation can be quite evident. For \( f(x) = \frac{ax+b}{cx-a} \):
  • The term \( ax + b \) can shift or tilt the graph depending on the values of \( a \) and \( b \), affecting the x-intercept.
  • The term \( cx - a \) in the denominator can move vertical asymptotes, which are lines the graph approaches but never crosses.
Understanding these transformations offers a deeper insight into analyzing function behavior visually and graphically. By manipulating constants and terms, you can predict how a graph's shape might change, a valuable skill for solving complex mathematical problems effectively.

Thus, interpreting transformations helps grasp the graphic nature of functions like \( f(x) = \frac{ax+b}{cx-a} \) and confirms insights such as graph symmetry, seen in our original exercise when considering the conditions provided.

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Most popular questions from this chapter

The number of hours of daylight on a given day at a given point on the Earth's surface depends on the latitude \(\lambda\) of the point, the angle \(\gamma\) through which the Earth has moved in its orbital plane during the time period from the vernal equinox (March \(21),\) and the angle of inclination \(\phi\) of the Earth's axis of rotation measured from ecliptic north \(\left(\phi \approx 23.45^{\circ}\right) .\) The number of hours of daylight \(h\) can be approximated by the formula $$ h=\left\\{\begin{array}{ll}{24,} & {D \geq 1} \\ {12+\frac{2}{15} \sin ^{-1} D,} & {|D|<1} \\ {0,} & {D \leq-1}\end{array}\right. $$ $$ \begin{array}{l}{\text { where }} \\ {\qquad D=\frac{\sin \phi \sin \gamma \tan \lambda}{\sqrt{1-\sin ^{2} \phi \sin ^{2} \gamma}}}\end{array} $$ and \(\sin ^{-1} D\) is in degree measure. Given that Fairbanks, Alaska, is located at a latitude of \(\lambda=65^{\circ} \mathrm{N}\) and also that \(\gamma=90^{\circ}\) on June 20 and \(\gamma=270^{\circ}\) on December \(20,\) approximate (a) the maximum number of daylight hours at Fairbanks to one decimal place (b) the minimum number of daylight hours at Fairbanks to one decimal place.

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Determine whether the statement is true or false. Explain your answer. A one-to-one function is invertible.

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