The temperature of a liquid during an experiment can be modeled by the function f(x)=3.8cos(πx20)+2.2 , where f(x) is the temperature in °C and x is the number of minutes into the experiment.
What is the lowest temperature the liquid reached during the experiment?
Round to the nearest tenth of a degree if needed.
Use 3.14 for π .
confused please help!!
since the lowest value of cos(x) is -1. the lowest temperature is
3.8(-1)+2.2 = -1.6
Sure! I can help you with that. To find the lowest temperature the liquid reached during the experiment, we need to identify the minimum value of the function.
The function given is f(x) = 3.8cos(πx/20) + 2.2, where x represents the number of minutes into the experiment.
To find the minimum value, we can consider the maximum value of the cosine function, since the cosine function can only range between -1 and 1. We need to find the maximum value of cos(πx/20).
The maximum value of cos(θ) is 1, which occurs when the angle θ is 0. Therefore, we need to find the value of x that makes πx/20 equal to 0.
To find that value, we set πx/20 = 0, and solve for x:
πx/20 = 0
Since any number multiplied by 0 equals 0, this equation is true for all x.
So, to find the lowest temperature, we substitute x = 0 into the original function:
f(0) = 3.8cos(π(0)/20) + 2.2
f(0) = 3.8cos(0) + 2.2
Since cos(0) = 1, we have:
f(0) = 3.8(1) + 2.2
f(0) = 3.8 + 2.2
f(0) = 6
Therefore, the lowest temperature the liquid reached during the experiment is 6°C.
To find the lowest temperature the liquid reached during the experiment, we need to understand the characteristics of the given function.
The function is \(f(x) = 3.8 \cos(\pi x/20) + 2.2\), where \(f(x)\) represents the temperature in degrees Celsius and \(x\) represents the number of minutes into the experiment.
In this function, \(\cos(\pi x/20)\) represents the oscillation or variation of the temperature. The amplitude of this oscillation is 3.8, meaning the temperature varies between 3.8 degrees above and below the average temperature. The average temperature is given by the constant term 2.2.
Since we want to find the lowest temperature, we need to identify the minimum value of the function.
To find the minimum, we need to determine the x-value that will cause \(\cos(\pi x/20)\) to be at its lowest point, which is when it is equal to -1.
Setting \(\cos(\pi x/20)\) equal to -1, we have:
\(-1 = \cos(\pi x/20)\)
To solve for \(x\), we can take the inverse cosine (cos^-1) of both sides:
\(\cos^{-1}(-1) = \cos^{-1}(\cos(\pi x/20))\)
Simplifying further, we have:
\(\pi x/20 = \pi + 2k\pi\) (where \(k\) is an integer)
Rearranging the equation, we get:
\(\pi x/20 = \pi(1 + 2k)\)
Now, isolate \(x\) by multiplying both sides by 20/\(\pi\):
\(x = 20(1 + 2k)\)
Since we are interested in the smallest \(x\)-value at which the temperature reaches its lowest point, we can set \(k\) to zero:
\(x = 20(1 + 2(0))\)
Simplifying further, we find:
\(x = 20\)
Now that we have the \(x\)-value, we can substitute it into the original function to find the lowest temperature:
\(f(x) = 3.8 \cos(\pi \cdot 20/20) + 2.2\)
Simplifying, we have:
\(f(x) = 3.8 \cos(\pi) + 2.2\)
Since \(\cos(\pi) = -1\), we get:
\(f(x) = -3.8 + 2.2\)
Finally, calculating the expression:
\(f(x) = -1.6\)
Therefore, the lowest temperature the liquid reached during the experiment is -1.6 degrees Celsius.