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Now that we’ve discussed the functions which correspond to horizontal lines, \(y = b\), we move to discussing the functions which can be represented by lines of the form \(y = mx + b\) where \(m \neq 0\). These functions are called linear functions and are described below.
As with Definition constantfunction, in Definition linearfunction, \(x\) is the independent variable, \(f\) is the function name, and both \(m\) and \(b\) are parameters. Notice that \(m\) is restricted by \(m \neq 0\) for if \(m = 0\) then the function \(f(x) = mx + b\) would reduce to the constant function \(f(x) = b\). The domain of linear functions, like that of constant functions, is specified as \((-\infty , \infty )\)
Recall that the form of the line \(y = mx + b\) is called the slope-intercept form of the line and the slope, \(m\), and the \(y\)-intercept \((0, b)\), are easily determined when the line is written this way. Likewise, the form of the function in Definition linearfunction, \(f(x) = mx + b\), is often called the slope-intercept form of a linear function.
The graph of a linear function is the graph of the line \(y = mx + b\). Lines are uniquely determined by two points, and two points of geometric interest are the axis intercepts. We’ve already reminded you of the \(y\)-intercept, \((0,b)\), which is obtained by setting \(x = 0\). Similarly, to find the \(x\)-intercept, we set \(y = 0\) and solve \(mx + b = 0\) for \(x\). We leave this to the reader in Exercise xinterceptoflinear. In addition to having special graphical significance, axis intercepts quite often play important roles in applications involving both linear and non-linear functions. For that reason, we take the time to define them here using function notation.
Suppose \(f\) is a function represented by the graph of \(y = f(x)\).
If \(0\) is in the domain of \(f\) then the point \((0, f(0))\) is the \(y\)-intercept of the graph of \(y = f(x)\).
That is, \((0,f(0))\) is where the graph meets the \(y\)-axis.
If \(0\) is in the range of \(f\) then the solutions to \(f(x) = 0\) are called the zeros of \(f\). If \(c\) is a zero of \(f\) then the point \((c,0)\) is an \(x\)-intercept of the graph of \(y = f(x)\).
That is, \((c,0)\) is where the graph meets the \(x\)-axis.
As is customary in this text, Definition interceptdefns uses the default independent variable \(x\), function name \(f\), and dependent variable \(y\), so these letters will change depending on the context. Also note that the ‘zeros’ of a function are the solutions to \(f(x) = 0\) - so they are real numbers. The \(x\)-intercepts are, on the other hand, points on the graph. As a quick example, consider \(f(x) = x-3\). The zeros of \(f\) are found by solving \(f(x) = 0\), or \(x-3=0\). We get one solution, \(x = 3\). Therefore, \(x=3\) is the zero of \(f\) that corresponds graphically to the \(x\)-intercept \((3,0)\).
We now turn our attention to slope. The role of slope, or more generally a ‘rate of change’, in Science and Mathematics cannot be overstated. As you may recall, or quickly read about on page ??, the slope of a line that has been graphed in the \(xy\)-plane is defined geometrically as follows:
where the capital Greek letter ‘\(\Delta \)’ denotes ‘change in.’ In this course, it is vital that we regard the slope of a linear function as a rate of change of function outputs to function inputs. That is, given the graph of a linear function \(y = f(x) = mx + b\):
What is important to note here is that for linear functions, the rate of change \(m\) is constant for all values in the domain. We’ll see the importance of this statement in the upcoming examples.
Geometrically, the sign of the slope has a profound impact on the graph of the line. Recall that if the slope \(m > 0\), the line rises as we read from left to right; if \(m<0\), the line falls as we read from left to right; if \(m=0\), we have a horizontal line and the graph plateaus. We define these notions more precisely for general functions in the following definition.
Let \(f\) be a function defined on an interval \(I\). Then \(f\) is said to be:
increasing on \(I\) if, whenever \(a < b\), then \(f(a) < f(b)\). (i.e., as inputs increase, outputs increase.)
NOTE: The graph of an increasing function rises as one moves from left to right.
decreasing on \(I\) if, whenever \(a < b\), then \(f(a) > f(b)\). (i.e., as inputs increase, outputs decrease.)
NOTE: The graph of a decreasing function falls as one moves from left to right.
constant on \(I\) if \(f(a) = f(b)\) for all \(a\), \(b\) in \(I\). (i.e., outputs don’t change with inputs.)
NOTE: The graph of a function that is constant over an interval is a horizontal line.
Again, as with Definition interceptdefns, Definition incdeccnstdefn applies to any function, not just linear and constant functions. Also, note that, like Definition absmaxmindefn, Definition incdeccnstdefn blurs the line between the function, \(f\), and its outputs, \(f(x)\), because the verbiage ‘\(f\) is increasing’ is really a statement about the outputs, \(f(x)\). Finally, when we ask ‘where’ a function is increasing, decreasing or constant, we are looking for an interval of inputs. We’ll have more to say about this in later sections, but for now, we summarize these ideas graphically below.
From the graphs above, we see that regardless if \(m>0\) or \(m<0\), the range of linear functions is \((-\infty , \infty )\). Therefore, linear functions have no maximum or minimum.
Solution.
To find \(C(0)\), we substitute \(0\) for \(x\) in the formula \(C(x)\) and obtain: \(C(0) = 80(0) + 150 = 150\). Given that \(x\) represents the number of PortaBoys produced and \(C(x)\) represents the cost to produce said PortaBoys, \(C(0) = 150\) means it costs \(\$150\) even if we don’t produce any PortaBoys at all. At first, this may not seem realistic, but that \(\$150\) is often called the fixed or start-up cost of the venture. Things like re-tooling equipment, leasing space, or any other ‘up front’ costs get lumped into the fixed cost. To find \(C(5)\), we substitute \(5\) for \(x\) in the formula \(C(x)\): \(C(5) = 80(5)+150 = 550\). This means it costs \(\$550\) to produce \(5\) PortaBoys for the local retailer. These two computations give us two points on the graph: \((0, C(0))\) and \((5, C(5))\). Along with the domain restriction \(x \geq 0\), we get:
The cost function \(C(x)= 80x + 150\) is in slope-intercept form so we recognize the slope as the coefficient of \(x\), \(m = 80\). With \(m > 0\), the function \(C\) is always increasing. This means that it costs more money to make more game systems. To interpret the slope as a rate of change, we note that the output, \(C(x)\), is the cost in dollars, while the input, \(x\), is the number of PortaBoys produced:
A couple of remarks about Example PortaBoyCost are in order. First, if \(x\) represents the number of PortaBoy game systems being produced, then \(x\) can really only take on whole number values. We will revisit this scenario in Section QuadraticFunctions where we will see how the approach presented here allows us to use more elegant techniques when analyzing the situation than a discrete data set would allow.
Second, once we know that the variable cost is \(\$80\) per PortaBoy, we can revisit a computation we did earlier in the example. We computed \(C(185) = 14950\) and needed to compute \(C(186)\). With \(186\) being just one more PortaBoy than \(185\), we can use the variable cost to get
which agrees with our earlier computation. If we wanted to find \(C(300)\), we could do something similar. Using \(300 - 185 = 115\), we can find \(C(300)\) as follows:
In general, we could rewrite \(C(x) = C(185) + 80(x - 115)\). This same reasoning shows that for any \(x_0\) in the domain of \(C\), we have \(C(x) = C(x_0) + 80(x - x_0)\) - a fact we invite the reader to verify.
Indeed, the computations above are at the heart of what it means to be a linear function: linear functions change at a constant rate known as the slope. To better see this algebraically, recall that given a point \((x_0, y_0)\) on a line along with the slope, \(m\), the point-slope form of the line is: \(y - y_0 = m(x - x_0)\). Rewriting, we get \(y = y_0 + m (x - x_0)\) and setting \(y = f(x)\) and \(y_0 = f(x_0)\) yields:
A few remarks are in order. First note that if the point \((x_0, f(x_0))\) is the \(y\)-intercept \((0, b)\), Equation linearfunctionpointslope immediately reduces to the slope-intercept form of the line: \( f(x) = f(x_0) + m (x - x_0) = b + m(x - 0) = mx + b,\) so you can use Equation linearfunctionpointslope exclusively from this point forward.
Second, if we write \(\Delta x = x - x_0\), then \(x = x_0 + \Delta x\) so we can rewrite Equation linearfunctionpointslope as follows:
In other words, changing the input by \(\Delta x\) results in changing the output by \(m \Delta x\). This tracks since
The fact that we can write \(\Delta \text {outputs} = m \Delta x\) for any choice of \(x_0\) is another way to see that for linear functions, the rate of change is constant. That is, the rate of change, \(m\), is the same for all values \(x_0\) in the domain. We’ll put Equation linearfunctionpointslope to good use in the next example.
We are asked to find a linear function \(p(x)\) ostensibly because the retailer has only two data points and two points are all that is needed to determine a unique line. We know that \(20\) PortaBoys were sold when the price was \(220\) dollars and double that, so \(40\) units, were sold when the price was \(190\) dollars. Using the language of function notation, these statements translate to \(p(20)=220\) and \(p(40)=190\), respectively. We first find the slope
To determine a reasonable domain for \(p\), we certainly require \(x \geq 0\), because we can’t sell a negative number of game systems. Next, we require \(p(x) \geq 0\), otherwise we’d be paying customers to ‘buy’ PortaBoys. Solving \(-1.5 x + 250 \geq 0\) results in \(x \leq 166.\overline {6}\). This shouldn’t be too surprising since our graph passes through the \(x\)-axis at \((166.\overline {6}, 0)\), going from positive \(y\)-values (hence, positive \(p(x)\) values) to negative \(y\) (hence negative \(p(x)\) values).
Given that \(x\) represents the number of PortaBoys sold, we need to choose to end the domain at either \(x = 166\) or \(x = 167\). We have that \(p(166) = 1 > 0\) but \(p(167) = -0.5 < 0\) so we settle on the domain \([0,166]\). Our final answer is \(p(x) = -1.5x + 250\) restricted to \(0 \leq x \leq 166\) which is graphed above on the right.
The function \(p\) in Example PortaBoyDemand is called the price-demand function (or, sometimes called more simply a ‘demand function’) because it returns the price \(p(x)\) associated with a certain demand \(x\) - that is, how many products will sell. These functions, along with cost functions like the one in Example PortaBoyCost, will be revisited in Example PortaBoyProfit.
Our next two examples focus on writing formulas for piecewise-defined functions, the second of which models a real-world situation.
Translating this to function notation means \(C(0) = 60\), \(C(1) = 60\), \(C(3.796) = 60\), and, in general, \(C(g) = 60\) for \(0 \leq g \leq 4\). What happens if we use more than \(4\) gigabytes? Let’s say we use \(6\) gigabytes. Per the plan, we are charged \(\$60\) for the first \(4\) and then \(\$5\) for each gigabyte over \(4\). Using \(6\) gigabytes means that we are \(2\) gigabytes over and our overage charge is \( (\$ 5)(2) = \$ 10\). The total cost is the base plus the overages or \(\$60 + \$10 = \$70\). In general, if \(g>4\), the expression \((g-4)\) computes the amount of data used over \(4\) gigabytes. Our base plus overage then comes to: \(60 + 5(g-4) = 5g+40\). Putting this together with our previous work, we get
We are graphing a line so we need to plot just one more point to determine the graph. From our work above, we know \(C(6) = 70\), so we use \((6,70)\) as our second point. Our graph is below. As with the graphs shown from Example timetempex1, we use ‘\(\asymp \)’ to denote a break in the vertical axis in order to better display the graph.