The function, f(x)=(3 x-7) x2 / 3, x ∈ R, is increasing for all x lying in :

Question

The function, f(x)=(3 x-7) x2 / 3, x ∈ R, is increasing for all x lying in : (A) (-∞, 0) ∪((3/7), ∞) (B) (-∞, 0) ∪((14/15), ∞) (C) (-∞, (14/15)) (D) (-∞,-(14/15)) ∪(0, ∞)

Tardigrade Admin · Accepted Answer

f(x)=(3 x-7) x2 / 3 ⇒ f(x)=3 x5 / 3-7 x2 / 3 ⇒ f prime(x)=5 x2 / 3-(14/3 x1 / 3) =(15 x-14/3 x1 / 3) >0 <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/mathematics/e936845faeaa57a2486c5b4422bcd163-.png" /> ∴ f(x)>0 ∀ x ∈(-∞, 0) ∪((14/15), ∞)

Q. The function, $f (x) = (3 x - 7) x^{2/3}, x \in R,$ is increasing for all $x$ lying in :

$(- \infty, 0) \cup (\frac{3}{7}, \infty)$

$(- \infty, 0) \cup (\frac{14}{15}, \infty)$

$(- \infty, \frac{14}{15})$

$(- \infty, - \frac{14}{15}) \cup (0, \infty)$

$f (x) = (3 x - 7) x^{2/3}$
$\Rightarrow f (x) = 3 x^{5/3} - 7 x^{2/3}$
$\Rightarrow f^{'} (x) = 5 x^{2/3} - \frac{14}{3 x ^{1/3}}$
$= \frac{15 x - 14}{3 x ^{1/3}} > 0$

$∴ f (x) > 0\forall x \in (- \infty, 0) \cup (\frac{14}{15}, \infty)$

Q. The function, f(x)=(3x−7)x2/3,x∈R, is increasing for all x lying in :

(−∞,0)∪(73​,∞)

(−∞,0)∪(1514​,∞)

(−∞,1514​)

(−∞,−1514​)∪(0,∞)