Find a formula for a function f that satisfies the following conditions.

Question

lim x→±∞ f(x) = 0,   lim x→0 f(x) = −∞,  f(3) = 0, lim x→6⁻ f(x) = ∞,   lim x→6⁺ f(x) = −∞

F(x)=___________

Answer 1

"lim x→±∞ f(x) = 0"

This is asking for a horizontal asymptote at x= 0, as x approaches infinity for both the negative and positive direction. The first hint that this might be a function from the parent function of (1/x)

"lim x→0 f(x) = −∞"

This is asking for a vertical asymptote at y=0, as x grows very small, y approaches negative infinity from both sides (left and right or positive and negative if x=0). This is yet another hint that this might be a (1/x) function. This must mean that x is on the denominator with an even degree.

 "f(3) = 0"

This is asking for the numerator to equal 0 when x =3. Thus, (x-3) could be on the numerator of the parent function (1/x) as a root to a polynomial.

" lim x→6⁻ f(x) = ∞"

As x approaches 6 from the right side, it approaches infinity.

"lim x→6⁺ f(x) = −∞"

As x approaches 6 from the left side, it approaches negative infinity. These 2 statements means there's a horizontal asymptote at 6 approaching from different sides. This must mean that it is an (x-6) is on the denominator to an odd power degree.

With all this information put together a function similar to the one below would satisfy these limits:

-(x-3(x-2)^2)/((x-6)^3*x^2))

As the degree in the numerator is lower than the degree in the denominator (1 to 5), x approaches 0 as it goes to infinity (can test using L' Hopitals). There is an even degree x on the denominator to cause x to approach negative infinity from both sides. The numerator resolves to an equation with a root of 3 that causes y=3 when x=3. And an odd degree for "x-6" to cause the different approaching sides of infinity as x approaches 6.