The function f defined by f(x) =x3 - 3x2 + 5x + 7 ,is

Question

The function f defined by f(x) =x3 - 3x2 + 5x + 7 ,is (A) increasing in R. (B) decreasing in R. (C) decreasing in (0, ∞) and increasing in (- ∞ , 0). (D) increasing in (0, ∞) and decreasing in (- ∞ , 0).

Tardigrade Admin · Accepted Answer

f(x)=x3-3 x2+5 x+7 f'(x)=3 x2-6 x+5>0 f'(x)=3 x2-6 x+5<0 x ∈ φ

Q. The function $f$ defined by $f (x) = x^{3} - 3 x^{2} + 5 x + 7$ ,is

increasing in R.

decreasing in R.

decreasing in $(0, \infty)$ and increasing in $(- \infty, 0)$ .

increasing in $(0, \infty)$ and decreasing in $(- \infty, 0)$ .

$f (x) = x^{3} - 3 x^{2} + 5 x + 7$
$f^{'} (x) = 3 x^{2} - 6 x + 5 > 0$
$f^{'} (x) = 3 x^{2} - 6 x + 5 < 0$
$x \in ϕ$

Q. The function f defined by f(x)=x3−3x2+5x+7 ,is

increasing in R.

decreasing in R.

decreasing in (0,∞) and increasing in (−∞,0).

increasing in (0,∞) and decreasing in (−∞,0).

f(x)=x3−3x2+5x+7 f′(x)=3x2−6x+5>0 f′(x)=3x2−6x+5<0 x∈ϕ

Q. The function $f$ defined by $f (x) = x^{3} - 3 x^{2} + 5 x + 7$ ,is

decreasing in $(0, \infty)$ and increasing in $(- \infty, 0)$ .

increasing in $(0, \infty)$ and decreasing in $(- \infty, 0)$ .

$f (x) = x^{3} - 3 x^{2} + 5 x + 7$
$f^{'} (x) = 3 x^{2} - 6 x + 5 > 0$
$f^{'} (x) = 3 x^{2} - 6 x + 5 < 0$
$x \in ϕ$