For the next two (2) items that follow: Consider the curves
\({\rm{f}}\left( {\rm{x}} \right) = {\rm{x}}\left| {\rm{x}} \right| - 1{\rm{\;and\;g}}\left( {\rm{x}} \right) = \left\{ {\begin{array}{*{20}{c}} {\frac{{3{\rm{x}}}}{2},{\rm{\;x}} > 0}\\ {2{\rm{x}},{\rm{\;x}} \le 0} \end{array}} \right.\) Where do the curves intersect?
At (2, 3) and (-1, -2)
Finding Intersection Points of Curves
The question asks for the points where two curves, defined by the functions \({\rm{f}}\left( {\rm{x}} \right)\) and \({\rm{g}}\left( {\rm{x}} \right)\), intersect. Intersection points occur where \({\rm{f}}\left( {\rm{x}} \right) = {\rm{g}}\left( {\rm{x}} \right)\). To find these points, we need to set the expressions for \({\rm{f}}\left( {\rm{x}} \right)\) and \({\rm{g}}\left( {\rm{x}} \right)\) equal to each other and solve for \(x\).
The function \({\rm{f}}\left( {\rm{x}} \right)\) is given by \({\rm{f}}\left( {\rm{x}} \right) = {\rm{x}}\left| {\rm{x}} \right| - 1\). This function's definition changes based on whether \(x\) is positive or negative due to the absolute value term \(|x|\).
- If \({\rm{x}} \ge 0\), then \(|x| = x\), so \({\rm{f}}\left( {\rm{x}} \right) = x(x) - 1 = {{\rm{x}}}^{\rm{2}} - 1\).
- If \({\rm{x}} < 0\), then \(|x| = -x\), so \({\rm{f}}\left( {\rm{x}} \right) = x(-x) - 1 = -{{\rm{x}}}^{\rm{2}} - 1\).
So, we can write \({\rm{f}}\left( {\rm{x}} \right)\) as a piecewise function:
$${\rm{f}}\left( {\rm{x}} \right) = \left\{ {\begin{array}{*{20}{c}} {{{\rm{x}}}^{\rm{2}} - 1,{\rm{\;x}} \ge 0}\\ { - {{\rm{x}}}^{\rm{2}} - 1,{\rm{\;x}} < 0} \end{array}} \right.$$
The function \({\rm{g}}\left( {\rm{x}} \right)\) is already given as a piecewise function:
$${\rm{g}}\left( {\rm{x}} \right) = \left\{ {\begin{array}{*{20}{c}} {\frac{{3{\rm{x}}}}{2},{\rm{\;x}} > 0}\\ {2{\rm{x}},{\rm{\;x}} \le 0} \end{array}} \right.$$
To find the intersection points, we need to solve \({\rm{f}}\left( {\rm{x}} \right) = {\rm{g}}\left( {\rm{x}} \right)\) by considering the different intervals for \(x\) defined by the piecewise functions.
Solving for Intersection Points - Case 1: \({\rm{x}} > 0\)
When \({\rm{x}} > 0\), we have \({\rm{f}}\left( {\rm{x}} \right) = {{\rm{x}}}^{\rm{2}} - 1\) and \({\rm{g}}\left( {\rm{x}} \right) = \frac{{3{\rm{x}}}}{2}\). We set them equal:
$${{\rm{x}}}^{\rm{2}} - 1 = \frac{{3{\rm{x}}}}{2}$$
Multiply the entire equation by 2 to remove the fraction:
$$2\left( {{{\rm{x}}}^{\rm{2}} - 1} \right) = 2\left( \frac{{3{\rm{x}}}}{2} \right)$$
$$2{{\rm{x}}}^{\rm{2}} - 2 = 3{\rm{x}}$$
Rearrange into a standard quadratic equation form \(ax^2 + bx + c = 0\):
$$2{{\rm{x}}}^{\rm{2}} - 3{\rm{x}} - 2 = 0$$
We can solve this quadratic equation by factoring or using the quadratic formula. Let's use factoring:
$$\left( {2{\rm{x}} + 1} \right)\left( {{\rm{x}} - 2} \right) = 0$$
This gives two possible solutions for \(x\):
- \(2x + 1 = 0 \implies 2x = -1 \implies x = -\frac{1}{2}\)
- \(x - 2 = 0 \implies x = 2\)
Since we are considering the case where \({\rm{x}} > 0\), the solution \(x = -\frac{1}{2}\) is not valid for this case. The only valid solution in this case is \(x = 2\).
Now, find the corresponding \(y\)-coordinate by substituting \(x = 2\) into either \(f(x)\) or \(g(x)\). Using \({\rm{g}}\left( {\rm{x}} \right) = \frac{{3{\rm{x}}}}{2}\) for \({\rm{x}} > 0\):
$${\rm{g}}\left( 2 \right) = \frac{{3 \times 2}}{2} = \frac{6}{2} = 3$$
Using \({\rm{f}}\left( {\rm{x}} \right) = {{\rm{x}}}^{\rm{2}} - 1\) for \({\rm{x}} \ge 0\):
$${\rm{f}}\left( 2 \right) = {{2}}^{\rm{2}} - 1 = 4 - 1 = 3$$
Both functions give \(y = 3\) when \(x = 2\). Thus, \((2, 3)\) is an intersection point.
Solving for Intersection Points - Case 2: \({\rm{x}} \le 0\)
When \({\rm{x}} \le 0\), we need to consider two sub-cases based on the definition of \(f(x)\): \(x=0\) and \(x < 0\).
Subcase 2a: \({\rm{x}} = 0\)
Let's check if the curves intersect at \(x=0\). For \({\rm{f}}\left( {\rm{x}} \right)\): \({\rm{f}}\left( 0 \right) = 0\left| 0 \right| - 1 = 0 - 1 = -1\). For \({\rm{g}}\left( {\rm{x}} \right)\) using the \({\rm{x}} \le 0\) part: \({\rm{g}}\left( 0 \right) = 2\left( 0 \right) = 0\).
Since \({\rm{f}}\left( 0 \right) \ne {\rm{g}}\left( 0 \right)\) (i.e., \(-1 \ne 0\)), the curves do not intersect at \(x=0\).
Subcase 2b: \({\rm{x}} < 0\)
When \({\rm{x}} < 0\), we have \({\rm{f}}\left( {\rm{x}} \right) = -{{\rm{x}}}^{\rm{2}} - 1\) and \({\rm{g}}\left( {\rm{x}} \right) = 2{\rm{x}}\). We set them equal:
$$-{{\rm{x}}}^{\rm{2}} - 1 = 2{\rm{x}}$$
Rearrange into a standard quadratic equation form:
$$0 = {{\rm{x}}}^{\rm{2}} + 2{\rm{x}} + 1$$
This equation is a perfect square trinomial:
$${\left( {{\rm{x}} + 1} \right)}^{\rm{2}} = 0$$
This gives one solution for \(x\):
- \(x + 1 = 0 \implies x = -1\)
Since we are considering the case where \({\rm{x}} < 0\), the solution \(x = -1\) is valid for this case.
Now, find the corresponding \(y\)-coordinate by substituting \(x = -1\) into either \(f(x)\) or \(g(x)\). Using \({\rm{g}}\left( {\rm{x}} \right) = 2{\rm{x}}\) for \({\rm{x}} \le 0\):
$${\rm{g}}\left( -1 \right) = 2\left( -1 \right) = -2$$
Using \({\rm{f}}\left( {\rm{x}} \right) = -{{\rm{x}}}^{\rm{2}} - 1\) for \({\rm{x}} < 0\):
$${\rm{f}}\left( -1 \right) = -{{\left( -1 \right)}}^{\rm{2}} - 1 = -\left( 1 \right) - 1 = -1 - 1 = -2$$
Both functions give \(y = -2\) when \(x = -1\). Thus, \((-1, -2)\) is another intersection point.
Summary of Intersection Points
By analyzing the different cases for \(x\), we found two points where the curves intersect:
- From Case 1 (\(x > 0\)): \((2, 3)\)
- From Case 2b (\(x < 0\)): \((-1, -2)\)
- From Subcase 2a (\(x = 0\)): No intersection.
Therefore, the curves intersect at \((2, 3)\) and \((-1, -2)\).
| Interval | f(x) | g(x) | Equation \(f(x) = g(x)\) | Solutions for x | Valid Solutions in Interval | Intersection Point(s) |
|---|---|---|---|---|---|---|
| \(x > 0\) | \(x^2 - 1\) | \(3x/2\) | \(x^2 - 1 = 3x/2\) | \(x = 2, x = -1/2\) | \(x = 2\) | (2, 3) |
| \(x = 0\) | \(0^2 - 1 = -1\) | \(2(0) = 0\) | \(-1 = 0\) | No solution | - | None |
| \(x < 0\) | \(-x^2 - 1\) | \(2x\) | \(-x^2 - 1 = 2x\) | \(x = -1\) | \(x = -1\) | (-1, -2) |
The intersection points are \((2, 3)\) and \((-1, -2)\).
Revision Table: Key Concepts
| Concept | Description | Relevance to Problem |
|---|---|---|
| Absolute Value Function | Defined as \(|x| = x\) for \(x \ge 0\) and \(|x| = -x\) for \(x < 0\). | Crucial for defining the piecewise nature of \(f(x)\). |
| Piecewise Function | A function defined by multiple sub-functions, each applying to a certain interval of the main function's domain. | Both \(f(x)\) and \(g(x)\) are piecewise functions, requiring analysis of different cases based on intervals. |
| Intersection of Curves | Points where the graphs of two functions meet. Found by setting the function expressions equal to each other (\(f(x) = g(x)\)) and solving. | The main goal of the problem is to find these points. |
| Solving Quadratic Equations | Finding the values of \(x\) that satisfy an equation of the form \(ax^2 + bx + c = 0\), typically using factoring or the quadratic formula. | Equations in each case simplify to quadratic equations that need to be solved. |
Additional Information: Graphing Piecewise Functions
Graphing piecewise functions helps visualize their behavior and potential intersection points. For \({\rm{f}}\left( {\rm{x}} \right) = {\rm{x}}\left| {\rm{x}} \right| - 1\):
- For \({\rm{x}} \ge 0\), it's a parabola \(y = x^2 - 1\), opening upwards, shifted down by 1. It starts at \((0, -1)\).
- For \({\rm{x}} < 0\), it's a downward-opening parabola \(y = -x^2 - 1\), shifted down by 1. It approaches \((0, -1)\) as \(x\) approaches 0 from the left. Note \(f(0) = -1\).
For \({\rm{g}}\left( {\rm{x}} \right) = \left\{ {\begin{array}{*{20}{c}} {\frac{{3{\rm{x}}}}{2},{\rm{\;x}} > 0}\\ {2{\rm{x}},{\rm{\;x}} \le 0} \end{array}} \right.\):
- For \({\rm{x}} > 0\), it's a line \(y = \frac{3}{2}x\) with a positive slope, starting from \((0, 0)\) (but not including it in this interval).
- For \({\rm{x}} \le 0\), it's a line \(y = 2x\) with a positive slope, passing through and including \((0, 0)\).
Graphing these would show the segments of the curves and visually confirm the intersection points we found algebraically at \((2, 3)\) and \((-1, -2)\). The point \((0, -1)\) for \(f(x)\) and \((0, 0)\) for \(g(x)\) confirm they do not meet at \(x=0\).
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\({\rm{f}}\left( {\rm{x}} \right) = {\rm{x}}\left| {\rm{x}} \right| - 1{\rm{\;and\;g}}\left( {\rm{x}} \right) = \left\{ {\begin{array}{*{20}{c}} {\frac{{3{\rm{x}}}}{2},{\rm{\;x}} > 0}\\ {2{\rm{x}},{\rm{\;x}} \le 0} \end{array}} \right.\)
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