1. The problem statement, all variables and given/known data Prove that 1/9 ≤ sqrt(66)-8 ≤ 1/8 2. Relevant equations Question comes with: Use the mean value theorem (f(b) − f(a)) / (b − a) 3. The attempt at a solution I can do "find c" problems fine with MVT, but I have no idea how to prove an inequality. I tried using sqrt(x) as a function, but got stuck. Don't know C. Thanks!
You must remember the interval where c is defined. c is in ]a,b[, where a = 64, and b = 66, now we can write 64 < c < 66. What can you do with the c in this inequality and the one in the MVT?
sqrt 64 = 8 and sqrt 66 appears in the inequality. Now it should be clear whether sqrt x should be the function.
Since [itex]\sqrt{64}= 8[/itex], it should be clear that they are using [itex]f(x)= \sqrt{x}[/itex] on an interval from 64 to 66: (f(66)- f(64))/(66-64)= [itex](\sqrt{66}- 8)/2[/itex] and that is equal to the derivative, [itex]1/2\sqrt{x}[/itex] for some value of x between 64 and 66. Obviously the largest that can be is [itex]1/2\sqrt{64}= 1/2(8)[/itex]. What is the smallest it can be?
Hi, I used sqrtx and got what you got, being 1/2sqrtx≤(sqrt(66)-8)/2≤1/2sqrtx which then equals 1/2sqrtx≤(sqrt(66)-8)/2≤1/16 then, times 2 to clear the 2 in the denominator, 1/sqrtx≤sqrt(66)-8≤1/8 This is 2/3 of the proof done, but how do you find the smallest? Logic says that it should be 81, so equals 1/9, which ends the proof, but I'm not sure how.
Do you remember the properties of the inequalities? The one you need to remember here is if x < y and y < z, therefore x < z.
c in the interval ]64, 66[, so we can write 64 < c < 66, but we want 81 in there, 64 < c < 66 < 81. How can you use the property to only have 64, c and 81 in the inequality?
So c<66, 66<81, then x<81....then 64<c<81, then value of x is between 64 and 81, and then you can plug in 64 and 81 on the two sides of the inequality 1/2sqrtx≤(sqrt(66)-8)/2≤1/2sqrtx giving you 1/9≤sqrt(66)-8≤1/8 Is that it?
OK, to reiterate: 64 < c < 66 (sqrts) Then, by using MVT, and 66=a 64=b, you get (sqrt66 -8)/2 Then find derivative of sqrtx, which is 1/2sqrtx. Then, using transitive property, prove that c<66, 66<81, then x<81....then 64<c<81 Then, 1/2sqrtx≤(sqrt(66)-8)/2≤1/2sqrtx So, 1/2sqrt81≤(sqrt(66)-8)/2≤1/2sqrt8 So, times 2 all sides Getting 1/sqrt81≤sqrt(66)-8≤1/sqrt64 Equals 1/9≤sqrt(66)-8≤1/8 And that is proven. Am I right?
How did you get this 1/2sqrtx ≤ (sqrt(66)-8)/2 ≤ 1/2sqrtx? I would use the MVT to have sqrt(66)-8 = 1/sqrt(c), then transform the inequality 66 < c < 81 to put the sqrt(66)-8 in it.
My bad. 64 < c < 66 (sqrts) Then, by using MVT, and 66=a 64=b, you get (sqrt66 -8)/2 Then find derivative of sqrtx, which is 1/2sqrtx. Then, using transitive property, prove that c<66, 66<81, then x<81....then 64<c<81 then (sqrt66 -8)/2 = 1/2sqrt(c) then sqrt(66)-8 = 1/sqrt(c) c=1/(sqrt66-8)² then 64 < 1/(sqrt66-8)² < 81 then 1/64 < (sqrt66-8)² < 1/81 then 1/8 < sqrt66-8 < 1/9 But that is reversed with what I want to prove 1/9 ≤ sqrt(66)-8 ≤ 1/8 ???? What Happened???
It seems right, I would've done it a slightly different way. With 64 < c < 81 I changed it to sqrt(64) < sqrt(c) < sqrt(81), 1/8 > sqrt(c) > 1/9, then with the MVT you have sqrt(66)-8 = 1/sqrt(c), then you substitute 1/8 > sqrt(66) - 8 > 1/9. But that's just me.