Course Notes for Calculus I

First I take the derivative.

Next I set the derivative to zero and solve.

The domain of this function is all real numbers. I split up this domain using the critical point to get the intervals. On each interval, I test the sign of the derivative.

Since the function changes from decreasing to increasing at the critical point, there must be a minimum at \(x = \frac{1}{7}\text{.}\)

First I take the derivative, using the quotient rule.

Next I set the derivative to zero and solve. I only need to set the numerator equal to zero.

The domain of this function is all real numbers. I split up this domain using the critical points to get the intervals. On each interval, I test the sign of the derivative.

At \(x = 1 - \sqrt{8}\text{,}\) the function changes from decreasing to inscreasing, so this point is a minimum. At \(x = 1 + \sqrt{8}\text{,}\) the function changes from increasing to decreasing, so this point is a maximum.

First I take the derivative, using the chain rule (I haven’t shown the steps here).

Next I set the derivative to zero and solve. Note that the exponential term can never be zero, so I am free to divide by it, leaving only the polynomial term remaining.

The domain of this function is all real numbers. I split up this domain using the critical point to get the intervals. On each interval, I test the sign of the derivative.

Since the function change from decreasing to increasing at the critical point, there must be a minimum at \(x = 0\text{.}\)

First I take the derivative. I used the chain rule (with \(u^{-1}\) as the outside function), but the quotient rule works just a well.

Next I set the derivative to zero and solve. Note I only need to set the numerator equal to zero.

This is not a root of the denominator, so this is a valid critical point. The domain of this function is all real numbers (since the denominator has no roots). We split up this domain using the critical point to get the intervals. On each interval, I test the sign of the derivative.

Since the function change from increasing to decreasing at the critical point, there must be a maximum at \(x = 3\text{.}\)

First I take the derivative.

Next I set the derivative to zero and solve.

The domain of this function is all real numbers, but I can assume \(t \geq \) for the model. I split up this domain using the critical point to get the intervals. On each interval, I test the sign of the derivative.

Since the function change from increasing to decreasing at the critical point, there must be a maximum at \(t = \frac{3}{2}\text{.}\) The value at \(t = \frac{3}{2}\) is \(-3 \left( \frac{3}{2} \right)^2 + 9 \frac{3}{2} + 7 = \frac{-27}{4} + \frac{27}{2} + 7 = \frac{27}{4} + \frac{28}{4} = \frac{55}{4}\text{.}\) The maximum height is \(\frac{55}{4}\) meters.

First I take the derivative, using the chain rule (I haven’t shown the steps).

Next I set the derivative to zero and solve. I only need to consider the numerator.

The domain of this function is all real numbers, since the quadratic in the square root is always positive. However, concentration cannot be negative, so in the model we will assume a domain of positive numbers. I split up this domain using the critical point to get the intervals. On each interval, I test the sign of the derivative.

Since the function change from decreasing to increasing at the critical point, there must be a minimum at \(c = 2\text{.}\) The value of the function at \(c=2\) is \(T(2) = 14 \sqrt{7 - 8 + 4} = 14 \sqrt{3} \doteq 24.25\text{.}\) The minimum bounding temperature if 24.25 degree celcius.

I start with the derivative.

Next I set the derivative to zero and solve.

The domain of this function is all numbers above 25. I split up this domain using the critical point to get the intervals. On each interval, I test the sign of the derivative.

Since the function change from decreasing to increasing at the critical point, there must be a minimum at \(m = \frac{115}{4} \text{.}\) The closest whole number to this is 29, so the ideal choice is 29 processors.

First we take the derivative.

Next I set the derivative to zero and solve.

The domain of this function is all real numbers, but the model only makes sense for positive numbers. I ignore the critical point at 0, since I am only interested in positive heights. I split up this domain using the critical point to get the intervals. On each interval, I test the sign of the derivative.

Since the function change from increasing to decreasing at the critical point, there must be a maximum at \(h = 16\text{.}\) Therefore, the best fuel efficiency happens at a height of 16 kilometers.

There is a possible maximum net benefit when the marginal cost and marginal benefit are equal. I calculate those two and set them equal to each other.

Then I look at intervals. Below this critical value, say at \(x=1\text{,}\) the marginal benefit is 2 and the marginal cost is \(\frac{1}{2}\text{.}\) Here the net benefit is increasing, since the marginal benefit exceeds the marginal cost. Above the critical value, say at \(x=4\text{,}\) the marginal cost is 2 and the marginal benefit is 1. The marginal cost exceeds the marginal benefit, so the net benefit is decreasing. I conclude that there is a maximum net benefit at approximatly \(x = 2.52\text{.}\)

For the break even point, I set the two functions each to each other and solve.

This is only one positive break even point, at approximately \(x = 6.34\text{,}\) so that is by default the largest. Figure 10.4.1 shows the behaviour of this marginal analysis model.

These cost and benefit functions seems at least somewhat reasonable. The benefit function curves down, representing a decreasing benefit with growing production. That behaviour is reasonable if the demand for a product eventually dries up. The cost function curves up, showing that producing more eventually ends up costing a great deal per unit. This might not be correct if economies of scale bring down the cost, but it might make sense if the skills or materials needed become more difficult to acquire.

There is a possile maximum net benefit when the marginal cost and marginal benefit are equal. I calculate those two and set them equal to each other.

Then I look at intervals. The marginal cost is always fixed, so I just need to look at the marginal benefit. When \(x \lt 89\text{,}\) say \(x = 27\text{,}\) the marginal benefit is \(\frac{2}{9}\text{,}\) which is larger than the marginal cost. When \(x \geq 90\text{,}\) say \(x = 125\text{,}\) then the marginal benefit is \(\frac{2}{25}\text{,}\) which is smaller than the marginal cost. Therefore, the net benefit is increasing before the critical point and decreasing after it, so there is a maximum.

For the break even point, I set the two functions each to each other and solve

A computer gives us the two approximate solution. Since the benefit function is increasing, the larger \(x\) value is the larger break even point. Figure 10.4.2 shows the behaviour of this marginal analysis model.

This cost function is linear, showing that producing more or less has no effect on the cost per unit. This is possible for some items or services. The benefit function curves down as in the previous activity, again likely representing that eventually the demand for the product or service becomes saturated.

There is a possile maximum net benefit when the marginal cost and marginal benefit are equal. I calculate those two and set them equal to each other. I need to work with two pieces, one for \(x \leq 2\) and another for \(x \gt 2\text{.}\) I will start with \(x \leq 2\text{.}\)

This solution is not in the range \(x \leq 2\text{,}\) so there are no critical points on this part of the domain. For all of this part of the domain, the marginal cost is bounded by \(\frac{4}{3}\text{,}\) which is always smaller than the marginal benefit, so the net benefit is always increasing. Now I’ll consider \(x \gt 2\text{.}\)

These can never be equal and the marginal cost is always larger than the marginal benefit. There are no critical points and the net benefit is always decreasing.

For the break even point, I set the two functions each to each other and solve. Again, I work in two pieces, first taking \(x \leq 2\text{.}\)

This point isn’t in this section of the domain, so it is not a break even point. It is also true that \(x=0\) satisfies this equality, but that isn’t an interesting break even point. Now let’s take \(x \gt 2\text{.}\)

There is a break even point. It is the only (non-zero) break even point and the benefit is always positive, so it must be the largest break even point. Figure 10.4.3 shows the behaviour of this marginal analysis model.

This cost function curves upwards, meaning that eventually the materials or skills needed become scare and it becomes more difficult to produce the item or service. The piecewise benefit function is more interesting. For the first while, benefit is linear, meaning that each unit produces the same benefit. However, the benefit suddenly becomes flat, showing that after a period of time, the demand is suddenly zero and no more benefit can be produces.