You are given the function f(x)=1/x^2+10. f(x) is concave up whenever: A. x^2−10 is negative. B. 3x^2−10 is positive. C. 3x^2+10 is negative. D. x^2+10 is negative.^ E. 3x^2−10 is negative. F. 3x^2+10 is positive. f(x) is concave down whenever: A. 3x^2−10 is negative. B. 3x^2+10 is positive. C. x^2−10 is negative. D. 3x^2+10 is negative. E. x^2+10 is negative. F. 3x^2−10 is positive. The points of inflection of f(x) are the same as: A. the zeros of 3x^2+10. B. the zeros of 3x^2−10. C. the zeros of x^2−10. D. the zeros x^2+10.

Accepted Solution

Answer:f(x) is concave up whenever:B. 3x²−10 is positivef(x) is concave down whenever:A. 3x²−10 is negativeThe points of inflection of f(x) are the same as:B. the zeros of 3x²−10Step-by-step explanation:Given the function f(x) = 1 / (x²+10)We can determine the concavity by finding the second derivative.If f"(x) > 0  ⇒  f(x) is concave upIff"(x) < 0  ⇒  f(x) is concave downThenf'(x) = (1 / (x²+10))' = -2x / (x²+10)²⇒  f"(x) = -2*(10-3x²) / (x²+10)³if   f"(x) = 0   ⇒  -2*(10-3x²) = 0    ⇒  3x²-10 = 0f(x) is concave up whenever 3x²−10 > 0 f(x) is concave down whenever 3x²−10 < 0 The points of inflection of f(x) are the same as the zeros of 3x²-10 it means that 3x²-10 = 0