Answer:
1 ≤ x < 7/4
Explanation:
The key to the problem is realizing the ordering of the function values is the same as the ordering of the function arguments.
This problem actually resolves to three inequalities. Two of them ensure the arguments of f( ) are within its domain. The third is the one that satisfies the inequality shown.
The three inequalities are ...
- -8 < 4x -3 < 4 . . . . . . first argument is in the domain
- -8 < 2 -x² < 4 . . . . . . second argument is in the domain
- 4x -3 ≥ 2 -x² . . . . . . function values have the right ordering
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-8 < 4x -3 < 4
Add 3 and divide by 4:
-5 < 4x < 7 . . . . . . . . . add 3
-5/4 < x < 7/4 . . . . . . . divide by 4
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-8 < 2 -x² < 4
Subtract 2 and multiply by -1.
-10 < -x² < 2 . . . . subtract 2
10 > x² > -2 . . . . multiply by -1 (reverses the inequality)
√10 > |x| . . . . . . take the square root
We note that the first inequality is more restrictive on the value of x.
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4x -3 ≥ 2 -x²
Since f(x) is an increasing function, function values have the same ordering as its arguments. That is, the requirement that
f(4x -3) ≥ f(2 -x²)
means that the arguments have the same order:
4x -3 ≥ 2 -x²
x² +4x -5 ≥ 0 . . . . . rearrange to standard form
(x +5)(x -1) ≥ 0 . . . . factor the quadratic
The product of these two factors changes sign when one of the factors changes sign. That is, the sign of the product is ...
- positive for x > 1
- negative for -5 < x < 1
- positive for x < -5
The values of x where the quadratic is zero are also in the solution set. Then the solution to this quadratic is ...
(x ≤ -5) U (x ≥ 1)
However, the above restrictions on the domain of x mean that the given inequality has the solution ...
1 ≤ x < 7/4
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Additional comment
The attachment shows a graphing calculator solution. We used f(x)=x as a simple "increasing" function for f(x). (The exact function doesn't matter.) Putting the restriction on its domain means the appropriate limits are imposed on the answer to the inequality we're trying to solve.
We also rearranged the given inequaltiy
f(4x -3) ≥ f(2 -x²)
to the form where the comparison is to zero:
f(4x -3) - f(2 -x²) ≥ 0
This is the form we actually used to arrive at the quadratic and its solution, above.
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As we noted initially, for an increasing function, the ordering of function values is the same as the ordering of the arguments. When the function is decreasing, the function values are the reverse of the ordering of the arguments. That is why we reverse the inequality when we multiply by -1.
2 > 1
-2 < -1
That is, for the decreasing function f(x) = -x, we have ...
The same thing holds for any decreasing function. One that we occasionally run across in algebra problems is the cosine function. In a suitable domain, larger angles have smaller cosines.