Recent content by takercena

Thanks for your reply. I really appreciate that. But I still don't get it as if you said there is a curve between -1.62 and .62, what is the turning point? It's can be 0 isn't it? As i mentioned before, the inner range is probably -2/5 because you can just factorize the y from (y-2)x^2 +...

Thanks, but after I get the x value, what should i do next? Can you tell me what is the value for horizontal asymptote?

yes that is the question but might be placed in wrong section. First, you try to solve the real linear factor for denominator which is not the exist here. 2nd I don't think vertical asymptote exist as first if you factorize it, it will always not equal to 0 both denominator and numerator.

It's not x = -2/5 but it's y = -2/5. And => is more 'than or equal' and otherwise. Let me show you my work First I rearrange this into like this from y = (2x^2 + 2x + 1)/(x^2 + x - 1 ) into y(x^2 + x - 1 ) - (2x^2 + 2x + 1) = 0. Then i use b^2 - 4ac => 0 and compare the coefficient from...

I am sorry, but can you help me how to find the vertical asymptote? I am total confuse now :(

1. Homework Statement Find the image set of function fx = (2x^2 + 2x + 1)/(x^2 + x - 1 ) The attempt at a solution 1. multiply the denominator with y and then assume x is real i get (y-2)^2 - 4(y - 2)(-y-1) => 0 5y^2 - 8y - 4 => 0 y = 2 and y = -2/5 But the answer is y => 2...

The limit will be 2. But that's not the only answer.

horizontal

Homework Statement Find the image set of function fx = (2x^2 + 2x + 1)/(x^2 + x - 1 ) The attempt at a solution 1. multiply the denominator with y and then assume x is real i get (y-2)^2 - 4(y - 2)(-y-1) => 0 5y^2 - 8y - 4 => 0 y = 2 and y = -2/5 But the answer is y => 2 and...

Oh i see. Thanks

Hello, please i still don't get it. I sketch 2 graph from the equation and at x = 0, there is a line there. But why f'(x) for e^|x| is undefined when x = 0?

So there are two answer right? -1/e^x and e^x. Why f'x is undefined at x = 0?

Homework Statement a)How to differentiate fx=e^|x|? b)Why when x = 0, f'(x) is undefined Homework Equations The Attempt at a Solution Is it d/dx e^|x| = e^|x|? I have no idea for this question.

Homework Statement a. Find the value of x such as fx < gx where fx = |2x -1| and gx = x(2-x) b. evaluate \displaystyle\int^1_0 [gx - fx]\,dx Homework Equations none The Attempt at a Solution For question a I make it into 2 equation to 2x-1 = 2x-x^2 and 1 - 2x = 2x - x^2. I solve it and find...

\left( \begin{array}{ccc} 0 & -5 & 3 \\ 3 & 14 & 0 \\ 0 & 5 & -3 \\ \end{array} \left| \begin{array}{c} 2 & -5 & -2 \\ \end{array} \right) By the way, I already get it. The answer is y/3 = (z-2)/3 = (x+(5/3))/-14. Thanks both of you