The temperature of a room during an experiment can be modeled by the function 23.7cos(3.14x/60)+18, where f(x) is the temperature °F and x is the number f hours into the experiment. What is the lowest temperature the room reached during the experiment?

41.7 is the lowest temperature the room reached during the experiment.

Step-by-step explanation:

Given : f(x)=23.7 cos(60πx)+18

Solution :

To find the lowest temperature we need to calculate the derivative of a given function

Then equate that derivative with 0 and calculate the value of x

Then find the double derivative and put the obtained value of x in double derivative

If the solution is positive then the temperature is lowest and if negative then the temperature is highest

Then put that value of x in original f(x)

Using the above steps solve the given function :

f(x)=23.7 cos(60πx)+18

Put the value of π =3.14

f(x)=23.7 cos(188.4x)+18

f'(x) =23.7 (- sin 188.4x)*188.4+0

⇒f'(x) = - 4465.08(sin 188.4x)

Now equate the derivative = 0

⇒f'(x) = - 4465.08(sin 188.4x) =0

⇒Sin 188.4x =0

Using calculator x = 0

NOW calculate f''(x)= - 4465.08(cos 188.4x)*188.4

= 841221.072(cos 188.4x)

Now put value of x =0 in double derivative

⇒ 841221.072(cos 188.4*0)

⇒841221.072(cos 0)

Since cos 0 =1

⇒841221.072>0

Positive hence the solution will be minimum

Now put this value of x in f(x)

⇒f(x)=23.7 cos(188.4*0)+18

⇒f(x)=23.7* cos 0+18

Since cos 0 =1

⇒f(x)=23.7+18

⇒f(x)=41.7

Thus 41.7 is the lowest temperature the room reached during the experiment.

(give brainlest please)