A) How do you prove that if 0(<or=)x(<or=)10, then 0(<or=)sqrt(x+1)(<or=)10?
B) So once that is found, then how can you prove that if 0(<or=)u(<or=)v(<or=)10, then 0(<or=)sqrt(u+1)(<or=)sqrt(v+1)(<or=)10?
How do you prove that if 0(<or=)x(<or=)10, then 0(<or=)sqrt(x+1)(<or=)10?
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does the square root increase (is the derivative positive) as x goes from 0 to 10 ?
If so the left side of the domain is minimum and the right side is maximum of the function and we only need to test the ends.
d (x+1)^.5 / dx = .5 /sqrt(x+1)
that is positive everywhere in the domain so all we have to prove is the end points.
0 </= x </= 10
if x = 0
sqrt x+1 = sqrt 1 = 1
if x = 10
sqrt x+1 = sqrt 11 = 3.32
so
1 </ sqrt(x+1) </= 3.32
A) To prove that if 0 (<=) x (<=) 10, then 0 (<=) sqrt(x+1) (<=) 10, we can follow a mathematical proof using logical deductions:
1. Start with the given inequality: 0 (<=) x (<=) 10.
2. Add 1 to both sides of the inequality: 1 (<=) x+1 (<=) 11.
3. Take the square root of all terms: sqrt(1) (<=) sqrt(x+1) (<=) sqrt(11).
4. Simplify: 1 (<=) sqrt(x+1) (<=) sqrt(11).
Therefore, we have proven that if 0 (<=) x (<=) 10, then 0 (<=) sqrt(x+1) (<=) 10.
B) Next, we need to prove that if 0 (<=) u (<=) v (<=) 10, then 0 (<=) sqrt(u+1) (<=) sqrt(v+1) (<=) 10. We can apply similar logical deductions:
1. Start with the given inequalities: 0 (<=) u (<=) v (<=) 10.
2. Add 1 to all terms: 1 (<=) u+1 (<=) v+1 (<=) 11.
3. Take the square root of all terms: sqrt(1) (<=) sqrt(u+1) (<=) sqrt(v+1) (<=) sqrt(11).
4. Simplify: 1 (<=) sqrt(u+1) (<=) sqrt(v+1) (<=) sqrt(11).
Therefore, we have proven that if 0 (<=) u (<=) v (<=) 10, then 0 (<=) sqrt(u+1) (<=) sqrt(v+1) (<=) 10.
By following these steps and making use of basic mathematical rules and properties, we have successfully proven the given inequalities.