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