Max Min by observation

[Sonal7]

\(\displaystyle y \ = \ \frac{4x}{1-2x} \ = \ \frac{4x\ - 2 \ + 2}{1-2x} \ = \ \frac{2(2x-1) \ + 2}{1-2x}\ =\ \frac{2 - 2(1-2x)}{1-2x}=\ 2 \cdot \ \frac{1 - (1-2x)}{1-2x}=\ (-2) \cdot \ \left[1 - \frac{1}{1-2x}\right]\)

Take time to do the algebra "correctly" - without that Calculus will be "far away"!

I already solved this but the detailed breakdown helps the general audience. but I am wondering about this now. Why did they say 4,0 is the max. I mean we know quadratic graphs but without doing dy/dx how did they know?

 

[Deleted member 4993]

I already solved this but the detailed breakdown helps the general audience. but I am wondering about this now. Why did they say 4,0 is the max. I mean we know quadratic graphs but without doing dy/dx how did they know?View attachment 21617

4x² is always positive - which is being "subtracted" from 4. No matter what x is - a positive number will be subtracted. Think about it...

 

[Sonal7]

4x² is always positive - which is being "subtracted" from 4. Think about it...

Its easy for you to say, its hard to visualise ?

 

[Steven G]

4x² is always positive - which is being "subtracted" from 4. No matter what x is - a positive number will be subtracted. Think about it...

4x^2 is NOT always positive! Sometimes it is 0. To get the max, you want to subtract the least from 1, which is when x^2=0. So the max 1-x^2 can be is 1. 4*1=4. So the max is 4.

 

[JeffM]

You know presumably that the x coordinate of the vertex of the parabola represented by

[MATH]ax^2 + bx + c[/MATH] occurs at [MATH]- \dfrac{b}{2a}.[/MATH]
The equation we have here is [MATH]y = 4 - 4x^2.[/MATH]
So, what are a and b in this case?

So the vertex is to be found at what value of x?

What then will the y value be at the vertex?