If the absolute maximum value of the function f(x)=(x2-2 x+7) e(4 x3-12 x2-180 x+31) in the interval [-3,0] is f(α), then:

Question

If the absolute maximum value of the function f(x)=(x2-2 x+7) e(4 x3-12 x2-180 x+31) in the interval [-3,0] is f(α), then: (A) α=0 (B) α=-3 (C) α ∈(-1,0) (D) α ∈(-3,-1)

Tardigrade Admin · Accepted Answer

f prime( x )= e (4 x 3-12 x 2-180 x +31)(12( x 2-2 x +7)( x +3)( x -5)+2( x -1)) text for x ∈[-3,0] ⇒ f prime( x ) < 0 f(x) is decreasing function on [-3, 0] The absolute maximum value of the function f(x) is at x=-3 ⇒ α=-3

Q. If the absolute maximum value of the function $f (x) = (x^{2} - 2 x + 7) e^{(4 x^{3} - 12 x^{2} - 180 x + 31)}$ in the interval $[- 3, 0]$ is $f (α)$ , then:

$α = 0$

$α = - 3$

$α \in (- 1, 0)$

$α \in (- 3, - 1)$

$f^{'} (x) = e^{(4 x^{3} - 12 x^{2} - 180 x + 31)} (12 (x^{2} - 2 x + 7) (x + 3) (x - 5) + 2 (x - 1))$
$for x \in [- 3, 0]$
$\Rightarrow f^{'} (x) < 0$
$f (x)$ is decreasing function on $[- 3, 0]$
The absolute maximum value of the function $f (x)$ is at $x = - 3$
$\Rightarrow α = - 3$

Q. If the absolute maximum value of the function f(x)=(x2−2x+7)e(4x3−12x2−180x+31) in the interval [−3,0] is f(α), then:

α=0

α=−3

α∈(−1,0)

α∈(−3,−1)