In which of the following intervals y=F(x) is concave up?

Question

Let F(x)= \( \int\limits_{1}^{x} \)  f(t)dt and f(t)=  \( \int\limits_{1}^{t^2} \)   \( \frac{1-u}{1-u^6} \)  du. In which of the following intervals y=F(x) is concave up?

a.(−1,0)

b.(1/2,3/2)

c.(−2,−1)

d.(0,1)

e.(−1,1)

Answer 1

Wenn man \( F(x) \) zweimal differenziert bekommt man

$$ F''(x) = 2x \frac{1-x^2}{1-x^{12}} $$

Jetzt prüfen wann \( F''(x) > 0 \) bzw. \( F''(x) < 0 \) gilt.