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
Therefore the greatest maximum y-value will be y=7
Answer:
Which function has the greatest maximum y-value?
f(x)
