In y = - 3 cos (x) - 6, what is the value for y when x is 0?

7

Step-by-step explanation:

We have to calculate the maximum and minimum of these functions.

f(x)=3 cos (2x)+4

1) we find the first derivative

f´(x)=-6 sin(2x)

2) We find those values that makes the first derivative equal to zero.

-6 sin(2x)=0

sin (2x)=0/(-6)

sin (2x)=0

2x=sin⁻¹ 0

2x=kπ

x=kπ/2               K=(...,-2,-1,0,1,2,...)

2) we find the second derivative and check if it has a maximum or minimum at x=kπ/2

f´´(x)=-12 cos (2x)

for example if k=0;

f´´(0)=-12 cos(2*0)=-12<0 ; because -12 is less than "0" ,it  has a maximum at x=kπ/2.

3) we find the maximum y-value:

if K=0; ⇒x=0

f(x)=3 cos (2x)+4

f(0)=3 cos (2*0)+4=3+4=7

The maximum y-value of f(x)=3 cos (2x)+4 is y=7.

g(x)

We can look at the graph of this function :

the maximum y-value is y=3.

h(x)

We can look at the table of this function;

 the maximum y-value of this function is y=-2

you didn't post any answer choices, so its not 100% clear what your teacher is looking for

however, log(15/7) is equivalent to log(15) - log(7)

i'm using the rule log(x/y) = log(x) - log(y)