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
for part b again the derivative is positive throughout the domain so if v is right of u then sqrt (1+v) > sqrt(1+u)
thank you!
Additionally,
C) They give a recursively defined sequence: a_1=0.3; a_(n+1)=sqrt((a_n)+1)for n>1
How do you find out the first five terms for it. then prove that this sequence converges. What is a specific theorem that will guarantee convergence, along with the algebraic results of parts A and B?
.3
sqrt 1.3 = 1.14
sqrt 2.14 = 1.46
sqrt 2.46 = 1.57
sqrt 2.57 = 1.60
hmmm, not getting bigger very fast.
let's see what happens to the derivative for large n
.5/sqrt(x+1)
ah ha, look at that. When n gets big, the derivative goes to zero. So the function stops changing.
But why would you look for the derivative to go to zero? Does it have to do anything with the theorem: If summation of a_n converges then limit_(n-->infinity) of a_n = 0. If so, what would the limit be approaching? 10 or infinity? But if not, then what theorem would we use? I know you explained about the larger n for the derivative, but I do not understand how that relates to one of the theorems.
But doesn't it converge to infinity and not 0?
we want it to converge to 0 right? But does it even converge if it goes to infinity, or is that divergence?
Do you do the limit on the derivative?
Or is there another way to prove convergence with a theorem of some sort?
A) To prove that if 0 (≤) x (≤) 10, then 0 (≤) √(x + 1) (≤) 10, we have to follow a few steps:
Step 1: Start with the given inequality, which is: 0 (≤) x (≤) 10.
Step 2: Add 1 to each side of the inequality: 1 (≤) x + 1 (≤) 11.
Step 3: Take the square root of each side of the inequality: √(1) (≤) √(x + 1) (≤) √(11).
Step 4: Simplify: 1 (≤) √(x + 1) (≤) √(11).
Step 5: Since 1 (≤) √(x + 1) and √(x + 1) (≤) √(11), we can combine the two inequalities and write it as: 1 (≤) √(x + 1) (≤) √(11).
Step 6: Finally, we use the given information that x (≤) 10 to conclude that √(x + 1) is also less than or equal to √(11). So, the final inequality is: 1 (≤) √(x + 1) (≤) √(11).
B) To prove that if 0 (≤) u (≤) v (≤) 10, then 0 (≤) √(u + 1) (≤) √(v + 1) (≤) 10, we can use the result from part A and follow these steps:
Step 1: Start with the given inequality: 0 (≤) u (≤) v (≤) 10.
Step 2: Apply the result from part A to the values of u and v separately. Using the result, we know that if 0 (≤) u (≤) 10, then 0 (≤) √(u + 1) (≤) 10. Similarly, if 0 (≤) v (≤) 10, then 0 (≤) √(v + 1) (≤) 10.
Step 3: Combine the two inequalities obtained in step 2 to form the final inequality: 0 (≤) √(u + 1) (≤) √(v + 1) (≤) 10.
This completes the proof that if 0 (≤) u (≤) v (≤) 10, then 0 (≤) √(u + 1) (≤) √(v + 1) (≤) 10.