Question

Compare the functions shown below:

f(x)

cosine graph with points at 0, negative 1 and pi over 2, 1 and pi, 3 and 3 pi over 2, 1 and 2 pi, negative 1
g(x)

x y
−6 −11
−5 −6
−4 −3
−3 −2
−2 −3
−1 −6
0 −11
h(x) = 2 cos x + 1


Which function has the greatest maximum y-value?

a. f(x)
b. g(x)
c. g(x) and h(x)
d. f(x) and h(x)

Answer 1

f(x) and h(x)

Explanation:

f(x) and h(x) have the greatest maximum y-value because if you graph out all the equations you can determine by graph which ones have the greatest y-value. When you graph the equations you find that f(x) and h(x) both have a greatest y-value of 3! While g(x) has a greatest y-value of -2.

*Please note that it is looking for the greatest y-VALUE not the greatest y-INTERCEPT.

Answer 2

We can find the local maxima and minima of any -continous- function by first finding where the slope is 0, as at this point maxima or minima exist.

Given an arbitrary function '
`f(x)`' we find the point of slope 0 by taking its first derivative and equaling to 0 ('
`(d)/(dx)f(x)=0`').

Lets, first, find the local extremes of the first function:

`f(x)=cos(x)`

`(d)/(dx) f(x)=(d)/(dx)cos(x)=-sin(x)=0`

`x=sin^(-1) (0)=0`

So our first function has a maxima at '
`x=0`' or at '
`y=f(0)=cos(0)=1`'.

Now we get the extremes for the second function:

`h(x)=2cos(x)+1`

`(d)/(dx) h(x)=(d)/(dx)[2cos(x)+1]=-2sin(x)=0`

`x=0`

So our second function has a maxima at '
`x=0`' or at '
`y=h(0)=2cos(0)+1=3`'.

Clearly, '
`h(0)=3\ \textgreater \ f(0)=1`', this means the second function '
`h(x)`' has the largest maxima -y value-.