The population of a colony of bacteria is modeled by the function p(x)=50(e^-x - e^-x^2)+10 , for 0 ≤ x, where population P is in thousands, x is in hours, and x = 0 corresponds to the moment of introduction of a certain chemical into the colony's environment. At which time below is the rate of population growth the greatest?
answer is 0.8
you want to find where p'(x) has a max
p' = 50(2x e^-x^2 - e^-x)
See http://www.wolframalpha.com/input/?i=2x+e^-x^2+-+e^-x
The rate of population growth is the greatest when the derivative of the population function is equal to zero. So let's find the derivative of the given function:
p'(x) = 50(-e^(-x) + 2xe^(-x^2))
To find when the rate of population growth is the greatest, we need to find the x-value that makes the derivative equal to zero. So, we set p'(x) = 0 and solve for x:
0 = 50(-e^(-x) + 2xe^(-x^2))
Now, I could give you a long, boring solution, but what's the point of that when we have a perfectly good joke to distract you?
Why was the math book sad?
Because it had too many problems!
Well, sorry for the distraction. Now, back to the problem-solving:
0 = -e^(-x) + 2xe^(-x^2)
Now, let's set e^(-x) as a common factor:
0 = e^(-x)(-1 + 2x(e^(-x^2)))
To make this equation true, either e^(-x) must be zero, or -1 + 2x(e^(-x^2)) must be zero. But, since e^(-x) is never zero, we can exclude that possibility.
So, we're left with:
-1 + 2x(e^(-x^2)) = 0
Now, don't be too alarmed by this equation. It may seem daunting, but you can crack it!
Why did the math book become friends with the history book?
Because they both had a lot of stories to tell!
Alright, you're still with me, good! Now, let's solve for x:
2x(e^(-x^2)) = 1
Dividing both sides by 2x, we get:
e^(-x^2) = 1/2x
Now, this equation might make you think back to your high school math class, but remember, you can do it!
Why was the equal sign so humble?
Because it knew it wasn't greater or less than anyone else!
Alright, time to keep pushing forward! Taking the natural logarithm (ln) of both sides, we have:
-ln(x^2) = ln(1/2x)
Simplifying further, we get:
-ln(x^2) = -ln(2x)
Now, dividing both sides by -1, we have:
ln(x^2) = ln(2x)
Exponentiating both sides with base e, we get:
e^(ln(x^2)) = e^(ln(2x))
Simplifying further, we have:
x^2 = 2x
Moving everything to one side, we have:
x^2 - 2x = 0
Factoring out an x, we get:
x(x - 2) = 0
So, either x = 0 or x - 2 = 0.
But wait, x = 0 doesn't make sense in this context because it means the population growth rate is being calculated at the moment of introduction of the chemical, which wouldn't make sense. So we can discard x = 0.
Therefore, x - 2 = 0, giving us x = 2.
So, the rate of population growth is the greatest at x = 2 hours.
I hope that joke-filled explanation made the problem a little more entertaining for you!
To find the time at which the rate of population growth is the greatest, we need to find the derivative of the population function and then solve for the value of x where the derivative is equal to zero.
Step 1: Find the derivative of the population function.
The derivative of the function p(x) with respect to x can be found using the chain rule and the exponential function properties. The derivative of e^x is e^x, and the derivative of e^(-x) is -e^(-x). The derivative of x^2 is 2x.
So, the derivative of p(x) = 50(e^-x - e^-x^2) + 10 with respect to x is:
p'(x) = 50(-e^-x + 2xe^-x^2)
Step 2: Set the derivative equal to zero and solve for x.
To find the time at which the rate of population growth is the greatest, we need to find the values of x where p'(x) = 0.
50(-e^-x + 2xe^-x^2) = 0
We can divide both sides by -e^-x to simplify the equation:
-1 + 2xe^-x^2 = 0
Rearranging the equation:
2xe^-x^2 = 1
Solving for x:
xe^-x^2 = 1/2
Since the exponential function e^(-x^2) is always positive, x must be positive as well.
So, to find the time at which the rate of population growth is the greatest, we need to solve for x > 0:
x = 1/2e^(x^2)
Unfortunately, this equation cannot be solved algebraically. We can use numerical methods, such as graphing or using a calculator or computer software, to approximate the value of x where the derivative is equal to zero.
Alternatively, you can also use optimization techniques to find the maximum or minimum of a function. In this case, you would need to find the critical points (where the derivative is equal to zero or undefined) and then use the second derivative test to determine whether each critical point is a maximum or minimum point. However, this method is more advanced and may require more mathematical knowledge.