Question

Compare the functions shown below: f(x) = 4 sin (2x − π) − 1 g(x) x y −1 6 0 1 1 −2 2 −3 3 −2 4 1 5 6 h(x) = (x − 2)2 + 4 Which function has the smallest minimum y-value?

Answer 1

Answer:

The function f(x) has smallest minimum y-value. The smallest minimum y-value is -5.

Step-by-step explanation:

The given function is

$f(x)=4\sin(2x-\pi)-1$

We know that the value of sinθ lies between -1 and 1.

$-1\leq \sin(2x-\pi)\leq 1$

Multiply 4 on each side.

$-4\leq 4\sin(2x-\pi)\leq 4$

Subtract 1 from each side.

$-5\leq 4\sin(2x-\pi)-1\leq 3$

$-5\leq f(x)\leq 3$

It means minimum value of f(x) is -5.

From the given table of g(x) it is noticed that the minimum y-value is -3 at x=2.

The given function is

$h(x)=(x-2)^2+4$               .... (1)

The vertex form of a parabola is

$P(x)=a(x-h)^2+k$            .... (2)

Where, (h,k) is vertex.

On comparing (1) and (2), we get

$a=1,h=2,k=4$

The vertex is (2,4). Since a=1>0, therefore it is an upward parabola and the vertex of an upward parabola is the point of minima.

For the function h(x) the minimum value of y is 4 at x=2.

The minimum y-value of function f(x), g(x) and h(x) are -5,-3 and 4 respectively.

Therefore function f(x) has smallest minimum y-value. The smallest minimum y-value is -5.

Answer 2

F(x) wins. The sine has an amplitude of 4, and 1 is subtracted, so the minimum y value is -5. g(x) is never smaller than -3, and h(x) is always positive!