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Question 92880This question is from textbook Algebra and Trigonometry
: If f(x)=x(x+3)(x-1), use interval notation to give all values of x where f(x)>0.
This question is from textbook Algebra and Trigonometry
Answer by bucky(2189)   (Show Source):
You can put this solution on YOUR website! Given:
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f(x)=x(x+3)(x-1)
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Suppose we replace f(x) by y so that the equation is a little more in the form of the way
we are used to thinking about graphs. If we do that the equation becomes:
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y = x(x+3)(x-1)
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Now suppose that a factor on the right side has a value of x such that the factor itself equals
zero. Having a factor equal to zero on the right side means that the entire right side becomes
zero because it involves a multiplication by zero. And if the right side equals zero, then
y equals zero.
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Now think to yourself, "If a point on a graph has a y value equal to zero, where must that
point be on the graph?" With a little thought you will convince yourself that the point
must be on the x-axis, because any point on the x-axis has a y-value of zero.
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With this in mind we can return to the equation:
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y = x(x+3)(x-1)
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and see that there are three points on the x-axis. The first point is at x = 0, because
when x is equal to zero, y also equals zero. Therefore, (0, 0) is a point on the graph.
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The second point is where x = -3 because when x = -3 the factor (x + 3) equals zero and
therefore y also equals zero. So the point (-3, 0) is on the graph.
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The final point on the x-axis is where x = 1, because when x is equal to +1, the factor (x - 1)
is equal to zero and therefore y = 0. So the point (1, 0) is on the graph.
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Now sketch a plot of these three points on a coordinate system ... on the x-axis put dots
at x = -3, at x = 0 (the origin), and at x = +1. The graph of the function must go through
all three of these points. Therefore, as your eyes move from left to right, the graph must
have one of two general shapes as noted below:
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(1) The graph could come downward through the point at x = -3, then between x = -3 and
x = 0 it would drop a little then curve and turn upward to go through the point where x = 0,
following which it would rise a little then curve downward and go back downward through the
point where x = 1 and continue downward. It can never return to cross the x-axis again.
Or the graph could have another shape:
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(2) The graph could come upward through the point at x = -3, then between x = -3 and
x = 0 it would rise a little then curve and turn downward to go through the point where x = 0,
following which it would sink a little then curve upward and go back up through the
point where x = 1 and continue upward. It can never return to cross the x-axis again.
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We have these two possibilities. The following graph shows them:
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The green graph depicts the graph of case (1) above and the red graph depicts the graph
described in case (2).
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It is fairly easy to find out which case applies. One way to do it is to pick a convenient value
for x between two of the points on the x-axis, substitute this value into the equation
and find out if y is positive or negative. For example, a convenient value of x might be
x = -1. It is between the x-axis crossings at x = -3 and x = 0. Substitute x = -1 into
the equation for y and you have:
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y = x(x + 3)(x - 1) = -1*(-1 + 3)*(-1 -1) = -1*(+2)*(-2) = +4
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This tells us that the point (-1, +4) is on the graph (y has a positive value when x = -1).
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This being the case, the green graph is not the correct one. Therefore, the red graph applies
and we are working with the graph:
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The problem asked you to set up inequalities to show the values of x where f(x) (or its
equivalent y) is greater than zero.
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Even without the exact graph we can now tell that f(x) or y is greater than zero when
x is between -3 and 0. So one answer is -3 < x < 0. And y is greater than zero when x
is greater than +1. So another answer is x > +1
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Those are the two regions on the x-axis where y (or f(x)) is greater than zero.
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I hope this helps you to understand the problem a little better and see how you can work it.
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