All Answers 356344 Answers

\(f=\dfrac{x-y}{x^4+y^4+6}\)

\(\text{intuition tells me }y=-x\\ \text{plugging this in we get}\\ f = \dfrac{2x}{2x^4+6} = \dfrac{1}{x^3+\frac 3 x}\\ \dfrac{df}{dx} = \dfrac{1}{\left(x^3+\frac 3 x\right)^2}\left(3x^2-\dfrac{3}{x^2}\right)\\ \dfrac{df}{dx}=0 \Rightarrow 3x^2=\dfrac{3}{x^2}\\ x^4=1,~x=\pm 1,~y=-x\\ x=-1,~y=1 \text{ will be a minimum}\\ x=1,~y=-1 \text{ will be a maximum}\)

\(\text{The maximum value is }\dfrac{2}{2+6}=\dfrac 1 4\)

This can all be verified by actually solving \(\nabla f = (0,0)\) for it's real roots.  These turn out to be as mentioned above.

I'm stumped....I feel like SOMEthing is missing,,....like is the frame equal width all the way around?  ...or maybe something relating the width (x) to the height (Y).........hmmmmmmm....

See here:

https://www.symbolab.com/solver/limit-calculator/%5Clim_%7Bx%5Cto0%7D%5Cleft(%5Cfrac%7Bsin%5Cleft(x%5E%7B9%7D%5Cright)%7D%7Bx%7D%5Cright)

Don't 'edit' your question after you post it and people (several) take their time to help you by responding!....Rather:  REPLY to your original post and explain that you made a mistake and the 'this' is what you meant........try that.....mucho better than ignoring everyone's efforts to help YOU !!!

I posted the wrong question and didn't want to edit it because last time I edited a question after I posted it someone got slightly upset. I figured just posting the actual question hours later would be better. 

OK....it is a little bit annoying when you don't check your PREVIOUS posts....it is a bit MORE annoying when you do not post your Q's correctly in the FIRST place....but rather, you IGNORE your previous post and then RE-POST  !!!!!      We are NOT here to do your homework....please respond to your PREVIOUS MISTAKEN post rather than ignoring YOU made a MISTAKE....GRRRRRRRRrrrrr......

Thank you!

Here is something to look at!

https://web2.0calc.com/questions/discriminant-problem 

First term  1 = r + r^2   →  r = 1 - r^2   →   r^2 = 1 - r

Sum  =     1 / (1 - r )   =      1  / [ 1 - (1 - r^2) ]      =   1 / r^2    

r - s  =

r - 1/r^2  =

[ r^3 - 1 ] / [  1 - r ] =

- [ 1 - r^3 ] / [ 1 - r ] =

- [ 1 - r ] [ 1 + r + r^2 ] / [ 1 - r ]  =

- [ 1 + r + r^2 ] =

- [ 1 + (r + r^2 ) ] =

- [ 1 + 1 ] =

-2

Is there an equation that goes with this?

If you meant 8t > 2, then all values of t > 1/4 will satisfy your equation.

Solve: r^2 + r - 1
r =1/2[sqrt(5) - 1]
s =2.6180339887
r - s = - 2

Looks good, GM !!!!!

We have a    parabola that opens upward......the vertex is (6, 1)

The y value of 1 is the minimum value on the graph......so....the range is   [ 1, infinity )

Rom could you check this please.

I do not know thise off by heart, I have to work them out every time

  \(If\;\;\; \theta=acos(x)\quad then\;\;\;find \; \frac{d\theta}{dx}\\~\\ x=cos\theta\\ \frac{dx}{d\theta}=-sin\theta\\ \)

Now we know that  cos(theta) =x

so I draw a right angled triangle  where this is true.

\(\text{adjacent=x     hypotenuse=1     opposite=} \sqrt{1-x^2}\)

\(\frac{dx}{d\theta}=-sin\theta\\ \frac{d\theta}{dx}=\frac{-1}{\sqrt{1-x^2}}\\ \)

So it seems to me that 

\(\displaystyle \int - \left[\dfrac{1}{\sqrt{1-x^2}}\right]~dx = \arccos(x) +c\)

Wolfram|alpha seems to agree with me.

Wait, so what is the derivative of arccos(x)?

contrapositive

conversion of a proposition from all A is B      to all not-B is not-A.     (Google definition)

Statement

If a mineral's structure is cubic, then it is not gold

A=mineral's structure is (cubit)

B=it is (not gold)

If A then B

contrapositive

A=mineral's structure is (cubit)

B=it is (not gold)

all not-B is not-A

If it is not (not gold)   then   it's mineral structure is not  (cubic.)

If it is      gold            then   it's mineral structure is not cubic.

On my computer , the + and equal sign are the same key....

Soooooooo MAYBE you meant   x^2+x = ³⁰

^{x^2}  +x -30 = 0

(x-5)(x+6) = 0      means x = 5  or  -6      (the OPPOSITE of what CPhill found for the other equation)  

I'm going to guess that you meant

x^2 - x =  30          subtract 30 from both sides

x^2 - x -30 =  0      factor

(x - 6) ( x + 5) = 0

Set both factors to 0  and solve for x and we get that    x = 6    or x = -5

Are you sure you're copying the question right?

Exactly the same

Post your answers if you want to and I'll check...

\((AB)^{-1} = B^{-1}A^{-1}\)

\((AB)^{-1}\begin{pmatrix}1\\2\end{pmatrix}=B^{-1}A^{-1}\begin{pmatrix}1\\2\end{pmatrix}\)

\(A\begin{pmatrix}2 \\-1\end{pmatrix}=\begin{pmatrix}1\\2\end{pmatrix} \Rightarrow A^{-1}\begin{pmatrix}1\\2\end{pmatrix} = \begin{pmatrix}2\\-1\end{pmatrix}\)

\((AB)^{-1}\begin{pmatrix}1\\2\end{pmatrix} = B^{-1}\begin{pmatrix}2\\-1\end{pmatrix}=\begin{pmatrix}1\\3\end{pmatrix}\\ \text{(for reasons identical to that of }A)\)

Correct, GM  !!!

Well done.....

Set each function to -3   and see if the x value you find falls within the domain of the given function

For instance

Set 

2x + 6 =  - 3      subtract  6 from both sides

2x = - 9         divide both sides by 2

x = -9/2  =  -4.5

Note that this value is  within the function's domain of   x ≤  - 4

Try this with the other two functions and see that you will find solutions that fall within their domains as well

If you need help....let me know

\(\displaystyle \int - \left[\dfrac{1}{\sqrt{1-x^2}}\right]~dx = -\arcsin(x) \text{, not }\arccos(x)\text{ for one}\)