If f (x) = x3 + bx2 + cx + d and 0

Question

If f (x) = x3 + bx2 + cx + d and 0 < b2 < c, then in (- ∞ , ∞) (A) f (x) is a strictly increasing function (B) f(x) has a local maxima (C) f (x) is a strictly decreasing function (D) f (x) is bounded

Tardigrade Admin · Accepted Answer

f(x) = x3 + bx2+cx+d, 0 < b2 < c f'(x) =3x2 + 2bx+c Discriminant = 4b2 - 12c = 4 (b2 -3c) < 0 ∴ f'(x) >0 ∀ x ∈ R ⇒ is strictly increasing f(x) ∀ x ∈ R

Q. If $f (x) = x^{3} + b x^{2} + c x + d$ and $0 < b^{2} < c$ , then in $(- \infty, \infty)$

f (x) is a strictly increasing function

f(x) has a local maxima

f (x) is a strictly decreasing function

f (x) is bounded

$f (x) = x^{3} + b x^{2} + c x + d, 0 < b^{2} < c$
$f^{'} (x) = 3 x^{2} + 2 b x + c$
Discriminant $= 4 b^{2} - 12 c = 4 (b^{2} - 3 c) < 0$
$∴ f^{'} (x) > 0\forall x \in R$
$\Rightarrow$ is strictly increasing $f (x) \forall x \in R$

Q. If f(x)=x3+bx2+cx+d and 0<b2<c, then in (−∞,∞)