Show that f is continuous on (−[infinity], [infinity]). f(x) = 1 − x2 if x ≤ 1 ln(x) if x > 1 On the interval (−[infinity], 1), f is ---Select--- function; therefore f is continuous on (−[infinity], 1). On the interval (1, [infinity]), f is ---Select--- function; therefore f is continuous on (1, [infinity]). At x = 1, lim x→1− f(x) = lim x→1− = , and lim x→1+ f(x) = lim x→1+ = , so lim x→1 f(x) = . Also, f(1) = . Thus, f is continuous at x = 1. We conclude that f is continuous on (−[infinity], [infinity]).

what "following statement"?

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answer: b. f(x) = (x + 2)^3 (x^2 - 7x + 3)^4

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polynomials always have as many roots as their order. this is a 4th order polynomial- hence 4 roots!

complex roots of polynomials always come in pairs. the quadratic equation has the "+/-" in that will alwaysgenerate m+ni amd m-ni, so 7-2i must be a root of this polynomial equation.

the order of the numerator and denominator are both 2. so the horizontal asymptote is y=a2/b2 =7

verticle asymptotes are the forbidden x-values. f(x) becomes undefined at x=-3 because this value will cause a zero denominator.

a graph of r(x) will have vertical asymptotes at ±1 because these values will make the denominator zero. so the domain of the function cannot include ±1.

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