|x-1| < 3-x^2
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If you sketch the two functions |x – 1| and 3 - x² you will note that
1. |x – 1| is “V-shaped” with the vertex of the V at x = 1
2. 3 - x² is a downward opening parabola with vertex at (0, 3)
3. There are two intersection points of these two functions:
a. one intersection is with –(x – 1) and 3 ‒ x²: (-1, 2)
i.e., -x + 1 = 3 ‒ x² or x² ‒ x – 2 = 0 or (x – 2)(x + 1) = 0 with solutions x = 2 or x = -1. The x = 2 is ignored as this leg of |x – 1| does not include x = 2
b. the other intersection is with x – 1 and 3 - x²: ((-1 + √17)/2, (-1 + √17)/2 – 1)
i.e., x – 1 = 3 ‒ x² or x² + x – 4 = 0. Solutions are via quadratic formula
x = (-1 ± √17)/2, but (-1 ‒ √17)/2 is ignored as this leg of |x – 1| does not include negative values for x.
Therefore the solution interval for |x – 1| < 3 ‒ x² is (-1, (-1 + √17)/2)
PS: I used Microsoft Excel to sketch the graphs. With it I can change the starting values of x and increments of the x values. This makes it easier to see the minute detail of the functions.
1. |x – 1| is “V-shaped” with the vertex of the V at x = 1
2. 3 - x² is a downward opening parabola with vertex at (0, 3)
3. There are two intersection points of these two functions:
a. one intersection is with –(x – 1) and 3 ‒ x²: (-1, 2)
i.e., -x + 1 = 3 ‒ x² or x² ‒ x – 2 = 0 or (x – 2)(x + 1) = 0 with solutions x = 2 or x = -1. The x = 2 is ignored as this leg of |x – 1| does not include x = 2
b. the other intersection is with x – 1 and 3 - x²: ((-1 + √17)/2, (-1 + √17)/2 – 1)
i.e., x – 1 = 3 ‒ x² or x² + x – 4 = 0. Solutions are via quadratic formula
x = (-1 ± √17)/2, but (-1 ‒ √17)/2 is ignored as this leg of |x – 1| does not include negative values for x.
Therefore the solution interval for |x – 1| < 3 ‒ x² is (-1, (-1 + √17)/2)
PS: I used Microsoft Excel to sketch the graphs. With it I can change the starting values of x and increments of the x values. This makes it easier to see the minute detail of the functions.
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First you need to redefine abs(x-1) as
x-1 ,if x ≥ 1
-x+1, if x>1
Case1: x ≥ 1 the inequality becomes
x-1< 3 - x^2 or
x^2 +x-4< 0
Use the quadratic formula to solve
x=-1±sqrt(17)/2
The two values approx are -2.56 and 1.56.the inequality is negative over the interval
(-2.56, 1.56)
Case2:x <1 the inequality becomes
-x+1<3 - x^2 or
x^2 -x -2<0
(x-2)(x+1) <0
The inequality is negative over the interval (-1,2)
Combining the solutions we found for both cases, we get
(-2.56, 2)
You can use the radical for instead of the decimals if you prefer.
x-1 ,if x ≥ 1
-x+1, if x>1
Case1: x ≥ 1 the inequality becomes
x-1< 3 - x^2 or
x^2 +x-4< 0
Use the quadratic formula to solve
x=-1±sqrt(17)/2
The two values approx are -2.56 and 1.56.the inequality is negative over the interval
(-2.56, 1.56)
Case2:x <1 the inequality becomes
-x+1<3 - x^2 or
x^2 -x -2<0
(x-2)(x+1) <0
The inequality is negative over the interval (-1,2)
Combining the solutions we found for both cases, we get
(-2.56, 2)
You can use the radical for instead of the decimals if you prefer.
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If you sketch a rough plot of y = |x - 1| and y = 3 - x² on the same set of axes, it is clear that the two curves will intersect in two places---one in quadrant I to the right of x = 1 and once in quadrant II.
Find the intersections:
3 - x² = |x - 1| = x - 1 for x > 1 ==> x² + x - 4 = 0 ==> x = (√(17) -1)/2.
3 - x² = |x - 1| = 1 - x for x < 1 ==> x² - x - 2 = 0 ==> x = -1.
The inequality is true for x between these values (where the graph of the quadratic is on top).
-1 < x < (√(17) - 1)/2.
Find the intersections:
3 - x² = |x - 1| = x - 1 for x > 1 ==> x² + x - 4 = 0 ==> x = (√(17) -1)/2.
3 - x² = |x - 1| = 1 - x for x < 1 ==> x² - x - 2 = 0 ==> x = -1.
The inequality is true for x between these values (where the graph of the quadratic is on top).
-1 < x < (√(17) - 1)/2.