Question

Given f(x)=-2(x-a)^-2(x-b), where a and b are positive real numbers such that a0?

Answer 1

Not too complicated. Multiply, simplify, find roots, find where the function is negative.\[ f(x)=-2(x-a)^2-2(x-b)=-2(x^2-2ax+a^2)-2x+2b\]\[=-2(x^2+(1-2a)x+(a^2-b))\]

Answer 2

Use one of your standard methods to find the roots (complete the square or quadratic formula). Your solution will be something of a mess, but when you get the roots, you know that the parts of the domain between the roots produce the positive part of the the range. (This is true since the function has a leading coefficient of -2, so it opens downward)

Answer 3

Oh sorry, what if y=-2(x-a)^2(x-b)

Answer 4

A product?

Answer 5

yes

Answer 6

In that case, the function is a cubic, with a negative coefficient. That means the end behavior is positive to the left, negative to the right. The two roots are x=a and x=b. The function does not change sign at the x=a (even multiplicity), but does at x=b. That means the function is strictly greater than zero from negative infinity to b, not including a. More elegantly\[f(x)>0\implies x \in (-\infty,a)\cup(a,b)\]

Answer 7

Got it?

Answer 8

thank you) mmm...honestly, not much. let me read again)