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=?

Expert Answers

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(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)>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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