How To Find The Slope Of A Piecewise Function

Piecewise Functions

A piecewise function is a role defined by 2 or more expressions, where each expression is associated with a unique interval of the role's domain.

The domain of a function is the set of all possible real input values, usually represented by x.

The range of a role is the gear up of all possible real output values, unremarkably represented past y .

Example one:

What is the domain of the function graphed below?

Solution:

The given part is a piecewise function, and the domain of a piecewise part is the set of all possible x -values.

It is seen that the graph has breaks, known as discontinuities, at x = -iii and x = one. These discontinuities do not affect the domain of this function because the piecewise part is however defined at each discontinuity.

The graph begins at x = -7. At that place is a closed circumvolve at 10 = -7, which indicates that the value is in the domain of the function.

The graph ends at x = 3. There is an open circumvolve at 10 = 3, which indicates that the value is not in the domain of the function.

Therefore, the domain of the function is { ten | -7 ≤ 10 < 3} .

The x -intercepts , or zeros, of a role are the points where the graph of the function touches or crosses the x -axis. When the graph of a function touches or crosses the x -axis, f(x) = 0.

The y -intercepts of a function are the points where the graph of the function touches or crosses the y -axis. When the graph of a function touches or crosses the y -centrality, ten = 0.

Instance two:

Discover the x - and y -intercepts of the post-obit piecewise role.

$f(x) \ = \ \left\{\begin{matrix}5, \hspace*{78} &\text{if } & -\infty & \lt & x & \leq & -2\\ \vspace*{9 \hspace*{-10}}\\-\frac{1}{2}x \ + \ 4, \hspace*{35} &\text{if } & -2 & \lt & x & \lt & 2\\\vspace*{9 \hspace*{-10}}\\3x \ - \ x^{2}, \hspace*{15} &\text{if } & \ \ 2 & \leq & x & \lt & \infty\end{matrix}\right.$

Solution:

To notice the ten -intercept, or aught, of the piecewise part, let f(x) = 0.

To solve the equation f(x) = 0, fix each expression in the piecewise part equal to zero. And so, solve for x . Later on solving for x , make certain that the solution(south) of each equation be in the corresponding domain.

Prepare the first expression equal to zero, and solve.

$5 \ \neq \ 0$

Since five cannot equal 0, there are no x -intercepts in the outset section of the domain.

Gear up the 2d expression equal to zero, and solve.

$\begin{array}{rclC40C40C35}-\frac{1}{2}x \ + \ 4 &=& 0\\-\frac{1}{2}x &=& -4\\x &=& 8\end{array}$

Even though the equation can exist solved, x = 8 is not in 2d section of the domain; therefore, there are no x -intercepts in the second section department of the domain.

Set up the 3rd expression equal to zero, and solve.

$\begin{array}{c450}\begin{array}{rclC35C35}3x \ - \ x^{2} &=& 0\\x\left(3 \ - \ x\right) &=& 0\\\end{array}\\\vspace*{9 \hspace*{-10}}\\\begin{array}{rclcrclC35C35}x &=& 0 &\hspace*{30}& 3 \ - \ x &=& 0\\&& && x &=& 3\end{array}\end{array}$

In this case, the equation yielded ii solutions:x = 0 and x = iii. Fifty-fifty though x = 0 is a solution of the equation, it is not in third department of the domain. Although, the solution x = three is in the third section of the domain. And then, there is an an ten -intercept at x = 3.

To notice the y -intercept of the piecewise function, let x = 0.

Determine the expression that corresponds to the department of the domain that contains x = 0. In this case, 10 = 0 is in the second section of the function's domain.

Evaluate the expression that corresponds to the 2d section of the domain at 10 = 0.

$-\frac{1}{2}(0) \ + \ 4 \ = \ 0 \ + \ 4 \ = \ 4$

And so, there is a y -intercept at y = 4.

The 10 -intercept of the given piecewise function is (3, 0) , and the y -intercept is (0, 4) .

Discontinuities of a function are points where the graph of a part has breaks or gaps.

Example iii:

Observe whatever discontinuities of the graph of the following piecewise part.

Solution:

Discontinuities occur in piecewise functions at the shared endpoints of the domain sections.

To determine if a shared endpoint is a signal of discontinuity in a piecewise office, determine the two sections of the domain that contain the endpoint. So, evaluate each associated expression at the endpoint.

If both associated expressions evaluated at the endpoint are equal, so the piecewise function does not have a discontinuity at the point.
If both associated expressions evaluated at the endpoint are not equal, and then the piecewise function does have a discontinuity at the indicate.

In the given piecewise role, in that location are two shared endpoints of the domain sections:10 = -2 and x = 2. So, discontinuities could occur in the graph of the piecewise function at either, or both, of these points.

The endpoint 10 = -two is associated with the beginning and second sections of the domain.

The first section of the domain is associated with the expression 5.

The second section of the domain is associated with the expression $-\frac{1}{2}$ 10 + 4. Evaluate the expression at x = -ii.

$-\frac{1}{2}(-2) \ + \ 4 \ = \ 1 \ + \ 4 \ = \ 5$

Since v = 5, there is non a discontinuity at x = -2.

The endpoint ten = 2 is associated with the 2d and third sections of the domain.

The second section of the domain is associated with the expression $-\frac{1}{2}$ x + 4. Evaluate the expression at ten = 2.

$-\frac{1}{2}(2) \ + \ 4 \ = \ -1 \ + \ 4 \ = \ 3$

The third department of the domain is associated with the expression 3 x - x ^two . Evaluate the expression at 10 = 2.

$3(2) \ - \ (2)^{2} \ = \ 6 \ - \ 4 \ = \ 2$

Since 3 ≠ 2, in that location is a discontinuity at x = 2 .

The piecewise function is graphed below, showing the discontinuity at x = 2.

The graph of a function is increasing if the value of y increases as the value of ten increases.

The graph of a function is decreasing if the value of y decreases every bit the value of ten increases.

The graph of a function is abiding if the value of y does not change equally the value of x increases.

Example iv:

Determine the interval on which the graph of the following function is abiding.

Solution:

When determining the intervals on which a role is increasing, decreasing, or staying constant, always read the graph of the function from the negative 10 direction (the left) to the positive x direction (the right).

The graph of a function is constant if the value of y does not change equally the value of x increases.

Observing the graph from left to right, information technology is seen that the but interval on which the the values of y do not alter as the values of x increment is -four ≤ x < ane.

Therefore, the interval on which the graph of the part is constant is -iv ≤ 10 < 1 .

The average rate of change is the ratio of the change in f(ten) to the change in ten . $\text{average rate of change} \ = \ \frac{\text{change in }f(x)}{\text{change in }x}$

Case 5:

What is the average rate of alter between x = -2 and x = iv in the post-obit piecewise function?

$f(x) \ = \ \left\{\begin{matrix}-x, \hspace*{68} &\text{if } & -\infty & \lt & x & \lt & -3\\ \vspace*{9 \hspace*{-10}}\\x \ + \ 6, \hspace*{25} &\text{if } & -3 & \leq & x & \leq & 0\\\vspace*{9 \hspace*{-10}}\\3x, \hspace*{50} &\text{if } & \ \ 0 & \lt & x & \lt & \infty\end{matrix}\right.$

Solution:

The average charge per unit of change between two points, x ₁ and x ₂ , of a piecewise function can be constitute past dividing the difference in the function values at those points past the difference between the two points.

$\text{average rate of change} \ = \ \frac{f\left(x_{2}\right) \ - \ f\left(x_{1}\right)}{x_{2} \ - \ x_{1}}$

It is given that ten _ane = -two and x _two = 4.

Since the function is a piecewise part, determine which section of the domain contains x ₁ and x ₂ and determine the expression associated with the department of the domain. And so, evaluate the associated expression at each point.

The signal x ₁ = -2 is in the second section of the domain and is associated with the expression x + 6. Evaluate the associated expression at x ₁ .

$\begin{array}{rclB35C35}f\left(x_{1}\right) \ = \ f(-2) &=& (-2) \ + \ 6\\&=& 4\end{array}$

The point ten _two = 4 is in the third section of the domain and is associated with the expression iii x . Evaluate the associated expression at x ₂ .

$\begin{array}{rclB35C35}f\left(x_{2}\right) \ = \ f(4) &=& 3(4)\\&=& 12\end{array}$

Calculate the average rate of modify.

$\begin{array}{rclC50C45C45C45C45}\text{average rate of change} &=& \frac{f\left(4\right) \ - \ f\left(-2\right)}{4 \ - \ (-2)}\\&=& \frac{12 \ - \ 4}{4 \ + \ 2}\\&=& \frac{8}{6}\\&=& \frac{4}{3}\end{array}$

The average rate of modify of the piecewise office betwixt x = -two and x = iv is $\frac{4}{3}$ .

An accented value function can exist represented by a piecewise function, with two domain sections. One section of the domain of the piecewise role will stand for the portion of the absolute value function with a negative slope, while the other section of the domain of the piecewise part will represent the portion of the absolute value function with a positive slope.

Simply equally an accented value office has characteristics, such as a vertex, axis of symmetry, and maximum/minimum, a piecewise function can possess these characteristics every bit well.

Remember that the graph of a piecewise function, which represents an absolute value function, is "5"-shaped. This "5"-shaped graph is symmetric about a line, known as the centrality of symmetry, and information technology can open up or downwardly.

If the graph of a piecewise function, which represents an absolute value function, opens upwardly, then the function has a minimum value at its vertex.

If the graph of a piecewise role, which represents an absolute value office, opens down, and then the function has a maximum value at its vertex.

Example 6:

What is a piecewise function that represents the following office?

$f(x) \ = \ \mid x \ - \ 5\mid \ + \ 2$

Solution:

To write an absolute value part as a piecewise part, determine the section of the domain where the accented value function has a positive slope and the department of the domain where the absolute value function has a negative slope.

To determine the department of the domain where the accented value office has a positive gradient, set up the expression in the absolute value confined greater than or equal to zero, and solve for x .

$\begin{array}{rclC35C35}x \ - \ 5 &\geq& 0\\x &\geq& 5\end{array}$

Now, make up one's mind the expression that tin can represent the accented value function, where 10 ≥ 5.

$\begin{array}{rclC35C35}f(x) &=& x \ - \ 5 \ + \ 2\\&=& x \ - \ 3\end{array}$

To make up one's mind the section of the domain where the absolute value function has a negative slope, set the expression in the absolute value bars less than naught, and solve for x .

$\begin{array}{rclC35C35}x \ - \ 5 &\lt& 0\\x &\lt& 5\end{array}$

Now, determine the expression that tin can stand for the absolute value function, where x < five.

$\begin{array}{rclC35C35C35}f(x) &=& -\left(x \ - \ 5\right) \ + \ 2\\&=& -x \ + \ 5 \ + \ 2\\&=& -x \ + \ 7\end{array}$

Therefore, the piecewise function that can represent the given absolute value function is as follows.

$f(x) \ = \ \left\{\begin{matrix}-x \ + \ 7, \hspace*{35} &\text{if } & -\infty & \lt & x & \lt & 5\\ \vspace*{9 \hspace*{-10}}\\x \ - \ 3, \hspace*{23} &\text{if } & 5 & \leq & x & \lt & \infty\\\end{matrix}\right.$

Example 7:

Determine the vertex and the axis of symmetry of the following piecewise function.

$f(x) \ = \ \left\{\begin{matrix}-x \ + \ 6, \hspace*{35} &\text{if } & -\infty & \lt & x & \lt & 4\\ \vspace*{9 \hspace*{-10}}\\x \ - \ 2, \hspace*{23} &\text{if } & 4 & \leq & x & \lt & \infty\\\end{matrix}\right.$

Solution:

Determine whether the function has any discontinuities.

The endpoint associated with both sections of the domain is x = iv.

The offset department of the domain is associated with the expression - ten + 6. Evaluate the expression at x = 4.

$-(4) \ + \ 6 \ = \ 2$

The second section of the domain is associated with the expression x - two. Evaluate the expression at x = 4.

$4 \ - \ 2 \ = \ 2$

Since 2 = 2, there is not a discontinuity at x = four. Although, since each expression yielded the value of 2, when evaluated at the endpoint of the domain, the value, x = 2, is known every bit the critical value of the piecewise function.

Check the slope on either side of the critical value. If at that place is a slope sign change (positive to negative or negative to positive) with the same constant, so the end point of the domain will be the vertex.

The slope of the first department is -i, and the slope of the second section is i.

Since in that location is a change in slope from negative to positive and there is no discontinuity, the vertex of the piecewise function occurs at 10 = 4.

So, the vertex of piecewise office is (iv, 2) .

The axis of symmetry is the vertical line that passes through the vertex.

So, the axis of symmetry is given by the equation x = iv .

Example eight:

Determine the minimum of the piecewise role given in example 7.

Solution:

The piecewise function given in case 7 is an accented value function. An absolute value function has a maximum or minimum value at its vertex.

The vertex of the piecewise function given in instance 7 is at(4, 2), so the minimum of the function is at (4, 2) .

Source: https://sites.google.com/site/math1eoctreview/home/eoct-topics/piecewise

Posted by: hernandezwinger.blogspot.com

How To Find The Slope Of A Piecewise Function

Piecewise Functions

Example one:

Solution:

Instance two:

Solution:

Example iii:

Solution:

Example iv:

Solution:

Case 5:

Solution:

Example 6:

Solution:

Example 7:

Solution:

Example eight:

Solution:

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