Question

Question 7(Multiple Choice Worth 1 points)

Answer 1

Answer:

Option A.

Step-by-step explanation:

The given function is

$f(x)=0.15x^2-6x+400$

Leading coefficient is positive, so it is an upward parabola. Vertex of an upward parabola is the point of minima.

Vertex of a parabola $f(x)=ax^2+bx+c$ is

$Vertex=\left(\dfrac{-b}{2a},f(\dfrac{-b}{2a})\right)$

In the given function a=0.15, b=-6 and x=400.

$-\dfrac{b}{2a}=-\dfrac{-6}{2(0.15)}=20$

At x=20,

$f(20)=0.15(20)^2-6(20)+400=340$

So, minimum value of f(x) is 340 at x=20.

From the given table it is clear that the minimum value of g(x) is 55 at x=70.

Since 55 < 340, therefore, g(x) has a lower minimum values, i.e, (70,55).

Hence, option A is correct.