1. ## Derivatives/Continuity

1.)Show that there is one real number x such that 2cosx = x^2. Justify your answer

2.)Let g(x) = x^(2/5)
a.) is g continuous at x=0? Use the definition of continuity to justify.
b.) is g differentiable at x=0? Use the definition of differentiability to justify.

2. (1) Let $\displaystyle f(x)=2\cos{x}-x^2$. Note that $\displaystyle f(0)>0$ and $\displaystyle f\left(\frac{\pi}{2}\right)<0$. Now invoke intermediate-value theorem.

(2)(a) Hint: Given $\displaystyle \epsilon>0$, take $\displaystyle \delta=\epsilon^{\frac{5}{2}}.$

(2)(b) Hint: Does $\displaystyle \lim_{x\,\to\,0}{x^{-\frac{3}{5}}}$ exist?

3. You can show the first question grahically

4. Originally Posted by JaneBennet
(1) Let $\displaystyle f(x)=2\cos{x}-x^2$. Note that $\displaystyle f(0)>0$ and $\displaystyle f\left(\frac{\pi}{2}\right)<0$. Now invoke intermediate-value theorem.

(2)(a) Hint: Given $\displaystyle \epsilon>0$, take $\displaystyle \delta=\epsilon^{\frac{5}{2}}.$

(2)(b) Hint: Does $\displaystyle \lim_{x\,\to\,0}{x^{-\frac{3}{5}}}$ exist?
in 2b.) why x^3/5?

5. Work it out and see.

6. Originally Posted by JaneBennet
Work it out and see.
yes, it approaches 0. just a little lost as to the reason -why- i am taking the limit to the 3/5 power

7. Hello,
Originally Posted by NotEinstein
yes, it approaches 0. just a little lost as to the reason -why- i am taking the limit to the 3/5 power
Because $\displaystyle \left(x^{\frac 25}\right)'=\frac 25 \cdot x^{-\frac 35}$

8. Originally Posted by Moo
$\displaystyle \left(x^{\frac 25}\right)'=\frac 25 \cdot x^{-\frac 35}$
No, don’t do that. We are supposed to be using the definition of a derivative here.

The derivative of $\displaystyle x^{\frac25}$ at 0 is $\displaystyle \lim_{x\,\to\,0}{\frac{(0+x)^{\frac25}-0^{\frac25}}{(0+x)-0}}.$

9. Originally Posted by JaneBennet
(1) Let $\displaystyle f(x)=2\cos{x}-x^2$. Note that $\displaystyle f(0)>0$ and $\displaystyle f\left(\frac{\pi}{2}\right)<0$. Now invoke intermediate-value theorem.
This shows that there is at least one root in $\displaystyle (0,\pi/2)$, we still need to show that there is only one real root (that is if there is), which can be done by showing that $\displaystyle f(x)$ is strictly decreasing for x positive. Since f(x) is symmetric there are infact two real roots.

RonL

10. Originally Posted by CaptainBlack
This shows that there is at least one root in $\displaystyle (0,\pi/2)$, we still need to show that there is only one real root.

RonL
Does the question say we have to do it?