MATH SOLVE

7 months ago

Q:
# 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 −60 −11h(x) = 2 cos x + 1Which 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)

Accepted Solution

A:

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 '[tex]f(x)[/tex]' we find the point of slope 0 by taking its first derivative and equaling to 0 ('[tex] \frac{d}{dx}f(x)=0 [/tex]').

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

[tex]f(x)=cos(x)[/tex]

[tex] \frac{d}{dx} f(x)=\frac{d}{dx}cos(x)=-sin(x)=0[/tex]

[tex]x=sin^{-1} (0)=0[/tex]

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

Now we get the extremes for the second function:

[tex]h(x)=2cos(x)+1[/tex]

[tex]\frac{d}{dx} h(x)=\frac{d}{dx}[2cos(x)+1]=-2sin(x)=0[/tex]

[tex]x=0[/tex]

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

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

Given an arbitrary function '[tex]f(x)[/tex]' we find the point of slope 0 by taking its first derivative and equaling to 0 ('[tex] \frac{d}{dx}f(x)=0 [/tex]').

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

[tex]f(x)=cos(x)[/tex]

[tex] \frac{d}{dx} f(x)=\frac{d}{dx}cos(x)=-sin(x)=0[/tex]

[tex]x=sin^{-1} (0)=0[/tex]

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

Now we get the extremes for the second function:

[tex]h(x)=2cos(x)+1[/tex]

[tex]\frac{d}{dx} h(x)=\frac{d}{dx}[2cos(x)+1]=-2sin(x)=0[/tex]

[tex]x=0[/tex]

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

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