# f(x)= (x^3)/((x^2)-1) is defined on the interval [-20,19] A) The function f(x)  has vertical asympototes at ? B) f(x) is concave up on the region ?. C) The inflection point for this function is at x=?

`f(x)=(x^3)/((x^2-1))`

(a) The function has vertical asymptotes if it is reduced (no common factors in numerator and denominator) and the denominator is zero.

Thus `x^2-1=0 => x=+-1` . The vertical asymptotes are at x=-1 and x=1.

(b) The function is concave up on an interval where the 2nd derivative is...

`f(x)=(x^3)/((x^2-1))`

(a) The function has vertical asymptotes if it is reduced (no common factors in numerator and denominator) and the denominator is zero.

Thus `x^2-1=0 => x=+-1` . The vertical asymptotes are at x=-1 and x=1.

(b) The function is concave up on an interval where the 2nd derivative is positive.

`f(x)=(x^3)/((x^2-1))`

`f'(x)=((x^2-1)(3x^2)-x^3(2x))/((x^2-1)^2)=(x^4-3x^2)/((x^2-1)^2)`

`f''(x)=((x^4-2x^2+1)(4x^3-6x)-[(x^4-3x^2)(4x^3-4x)])/((x^2-1)^4)`

`=(2x^5-4x^3-6x)/((x^2-1)^4)`

`=(2x(x^2-1)(x^2+3))/((x^2-1)^4)`

`=(2x(x^2+3))/((x^2-1)^3)`

`f''(x)>0` on (-1,0) and `(1,oo)` so the given function is concave up on (-1,0) and (1,19)

(3) The inflection point occurs when `f''(x)=0` and the sign of the second derivative changes parity.

The second derivative is zero at x=0 which is the inflection point.

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