1. ## Calculus ProofS Help

Hey, I need help on a couple of problems in my proof class. If anyone can help, it would be highly appreciated, Thank you!

1. Prove that S = { n−1 |n ∈ N} is bounded above and that its supremumn
is equal to 1.
2.Use the Intermediate Value Theorem to show that the polynomial x4 +
x3 − 9 has at least two real roots.
3.Use the Mean Value Theorem to prove that |sin(b) − sin(a)| ≤ |b − a|
for all a, b ∈ R.

2. Originally Posted by Swamifez
2.Use the Intermediate Value Theorem to show that the polynomial x4 +
x3 − 9 has at least two real roots.
Let $\displaystyle f(x) = x^4+x^3-9$.

Note that for large, negative x f(x) takes a positive value.
f(0)= -9 is negative. Thus there is at least one root between a large negative x and x = 0.
For large, positive x f(x) takes a positive value again. Thus there is at least one root between x = 0 and large a large positive x.

Thus there are at least 2 real roots.

-Dan

3. Originally Posted by Swamifez
1. Prove that S = { n−1 |n ∈ N} is bounded above and that its supremumn
is equal to 1.
N = {1, 2, 3,...} yes? Then S = {0, 1, 2, ...}. This set is NOT bounded above!

-Dan

Ah wait! Are you defining $\displaystyle S = \{ n^{-1} | n \in N \}$?

Then show that for $\displaystyle p < q$ $\displaystyle p^{-1} > q^{-1}$ ($\displaystyle p, q \in N$). So the largest value in S is going to be for the smallest value in N: n = 1. Thus max(S) = 1. Since S has a maximum element sup(S) = max(S) (I believe. Someone please correct me if I'm wrong here!)

-Dan

4. Originally Posted by Swamifez
3.Use the Mean Value Theorem to prove that |sin(b) − sin(a)| ≤ |b − a|
for all a, b ∈ R.
If $\displaystyle b=a$ then the result is trivial (I love saying that).

If $\displaystyle b\not = a$ then without loss of generality (I love saying that too) you can assume $\displaystyle b>a$
Then, this $\displaystyle [a,b]$ is a closed interval.
Consider the function, $\displaystyle f(x)=\sin x$ defined on this interval. The function is continous on $\displaystyle [a,b]$ and differenciable on $\displaystyle (a,b)$ this satisfies the conditions of Lagrange's Mean Value Theorem. Thus, there exists a $\displaystyle c\in (a,b)$ such as,
$\displaystyle \frac{\sin b-\sin a}{b-a}=f'(c)=cos c$
Thus,
$\displaystyle \left| \frac{\sin b-\sin a}{b-a} \right|=|\cos c|\leq 1$
Thus,
$\displaystyle |\sin b-\sin a|\leq |b-a|$