Melody

Hi Pro35hp and MaxWong,

I think that you will both find this interesting.   It ties in nicely with Alan's graph as well.     

If you do not understand then please ask questions :)

\(\frac{x^2 + x + 3}{2x^2 + x - 6} \ge 0\\ \frac{x^2 + x + 3}{2x^2 + 4x-3x - 6} \ge 0\\ \frac{x^2 + x + 3}{2x(x+2)-3(x+2)} \ge 0\\ \frac{x^2 + x + 3}{(2x-3)(x+2)} \ge 0\\~\\ \mbox{Now I will consider the numerator and put it equal y}\quad y=x^2+x+3\\ \mbox{The discriminant is }1^2-12=-11 \qquad \mbox{ This means there are no real roots}\\ \mbox{Since the leading coefficiant is positive that means that }\quad y=x^2+x+3 \quad \mbox{is a concave up parabola}\\ \mbox{This means that for all real values of x the numerator will be positive}\ so\\ \mbox{If the denominator is > or = zero the statement will be true.}\\ True\;\;when\;\;\\ (2x-3)(x+2)\ge0\\ \mbox{If I graph }y=(2x-3)(x+2) \mbox{then this will be true when y is on or above the x axis, that is when } y \ge0\\~\\ \mbox{This is a concave up parabola so it will be true at the two ends}\\ \mbox{roots are } x=3/2\;\;\;and\;\;\ x=-2\\ \mbox{So the original statement will be true when }x\le-2\quad and \;\;when\;\; x\ge 1.5 \)

If y=arctan x, show that dy/dx=1/1+x squared, using the inverse of trigonometric functions. 

I am not sure if this was the intended method - I always do these from scratch.

\(y=atan(x)\\ x=tan(y)\\ \frac{dx}{dy}=sec^2(y)\\ \frac{dy}{dx}=\frac{1}{sec^2(y)}\\ \frac{dy}{dx}=cos^2(y)\\~\\ Given\;\; tan(y)=\frac{x}{1}\;\;then\;\;\\ If \;\;opp=x\;\;and\;\;adj=1\;\;then\;\;hyp=\sqrt{1+x^2} \;\;so\\ cos^2(y)=(\frac{1}{\sqrt{1+x^2}})^2=\frac{1}{1+x^2}\\~\\ \frac{dy}{dx}=\frac{1}{1+x^2}\\\)

f(x+1)=x.f(x)       f(5)=72      f(2)=?

\(f(5)=4*f(4)\\ 72=4*f(4)\\ f(4)=18\\~\\ f(4)=3*f(3)\\ 18=3*f(3)\\ f(3)=6\\~\\ f(3)=2*f(2)\\ 6=2*f(2)\\ f(2)=3\)

Try these steps and see how you go :)

You will learn better if you do quite a bit of it yourself.

You can post your answers (or troubles and someone will look at them for you :)

1) find the midpoint of AB

2) Find the gradient of AB

3)Determine the gradient of the perpendicular to AB

4) use the point gradient formula to determine the equation of the line.

Well we were not hard to find.  :)

We didn't go anywhere  LOL

Anyway it is good to see you back again :)

Hi Jillybean :)

Yes this is correct :)

6⁵

is entered as 6^5

 what is the square root symbol x=6 ? someone please help me

\(\sqrt{x}=6\\ (\sqrt{x})^2=6^2\\ x=36 \)

what is 6 and 5/8 times pi

(6+5/8)*pi = 20.8130513300323802