The intervals of increase of f(x) defined by f(x)=∫ limits-1x(t2+2 t)(t2-1) d t is equal to

Question

The intervals of increase of f(x) defined by f(x)=∫ limits-1x(t2+2 t)(t2-1) d t is equal to (A) (-∞, (-3/2)) ∪(0,3) ∪(10, ∞) (B) (-∞,-2) ∪(-1,0) ∪(1, ∞) (C) (-∞,-2) ∪((-1/2), (1/2)) ∪(4, ∞) (D) (-∞,-2) ∪((-3/4), (1/4)) ∪(1, ∞)

Tardigrade Admin · Accepted Answer

We have mathrmf prime( mathrmx)= mathrmx( mathrmx+2)( mathrmx+1)( mathrmx-1)

Q. The intervals of increase of $f (x)$ defined by $f (x) = - 1 \int x (t^{2} + 2 t) (t^{2} - 1) d t$ is equal to

$(- \infty, \frac{- 3}{2}) \cup (0, 3) \cup (10, \infty)$

$(- \infty, - 2) \cup (- 1, 0) \cup (1, \infty)$

$(- \infty, - 2) \cup (\frac{- 1}{2}, \frac{1}{2}) \cup (4, \infty)$

$(- \infty, - 2) \cup (\frac{- 3}{4}, \frac{1}{4}) \cup (1, \infty)$

We have $f^{'} (x) = x (x + 2) (x + 1) (x - 1)$

Q. The intervals of increase of f(x) defined by f(x)=−1∫x​(t2+2t)(t2−1)dt is equal to

(−∞,2−3​)∪(0,3)∪(10,∞)

(−∞,−2)∪(−1,0)∪(1,∞)

(−∞,−2)∪(2−1​,21​)∪(4,∞)

(−∞,−2)∪(4−3​,41​)∪(1,∞)