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# Rules and Theorems About Calculus

• Category: Science
• Subcategory: Math
• Topic: Calculus
• Pages: 4
• Words: 1949
• Published: 26 April 2019 Download Print

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121 writers online The wonderful world of math is strange yet beautiful and simple yet complex. As a kid we’re taught what numbers are, the numeric value they have, and how to use them. Then all of a sudden letters start showing up alongside the numbers and we’re told this is called algebra, an advanced form of math. Suddenly we feel like kings of the world, we know algebra! A few years later, we enter calculus and the definition of what we call math is no longer right. Now we’re adding, subtracting, multiplying, and dividing in our heads, plotting imaginary numbers, and graphing things in three dimensions. Sounds difficult right? But in reality it’s not, it’s actually quite simple.

Two main topics of calculus are the integral and the derivative. The derivative is the instantaneous rate of change of a function at a specific point. Another way to think about the derivative is how is the graph changing as x increases or decreases? The integral, or antiderivative as it is commonly called, is the opposite because it can be thought of as undoing the differentiation. Because the two are so closely related, many real-world connections can be made. If given a velocity graph, as shown in figure 1, the integral of the graph would represent the distance traveled or the position. Figure 2 shows the corresponding distance graph. Given the same velocity graph, the derivative of the graph would represent the acceleration as seen in figure 3.

The units for the distance graph are meters because the data displayed represents how far the object moved. The derivative of the distance graph, velocity graph, has units of meters per second because it represents how fast the object’s position is moving. The derivative of the velocity graph, and consequently the second derivative of the distance graph, is the acceleration graph. The acceleration graph as units of meters per second per second because it represents how fast the velocity is changing. A second example could be area. The integral of the area is volume and the derivative of area is length.

Often times a teacher will give a student a problem that says “given the graph of function f(x), draw the derivative, f’(x).” Or sometimes it might even be “given the graph of the derivative f’(x), draw in the original function f(x).” At first this may sound intimidating, but it’s really not as long as the a few key ideas are known. These key ideas are: critical points, sign changes of the first and second derivative, concavity and points of inflection, and maximum and minimums of the function. A critical point is a maximum or a minimum, it is a point where the derivative equals zero, and it is also the endpoints. Figure 4 shows the graph of function f(x) on the closed interval [a,e]. This graph has 5 critical points: a is an endpoint, b is a maximum, c is a minimum, d is a maximum and e is the other endpoint. If one were to draw the derivative graph of this, it would cross the x-axis, f’(x)=0, at each one of these points. A second key point is knowing what the sign changes of the first and second derivative graphs mean to the original function. In a derivative graph, the graph is above the x-axis the graph of the original function is increasing, and when the graph of a derivative is below the x-axis the graph of the original function is decreasing. Figure 5 shows the original, f(x), and the derivative, f’(x), on the approximate interval of -2

Another key idea is using the first and second derivatives to find points of inflection based on concavity. A point of inflection occurs when the second derivative, f’’(x), equals zero. An easy way to distinguish points of inflections using the original graph is a change in concavity. If a graph goes from concave up to concae down, of vice verse, a point of inflection occurs. If the first derivative graph goes from increasing to decreasing or vice versa then a point of inflection occcurs. Figure 6 shows a graph of a function approximately x=0 to x=3. The graph is concave down from x=0 to approximately x=1.4 and concave up from approximately x=1.4 to x=3, this means a point of inflection occurs at x=1.4.

In a graph, the global maximim and minimum are the absolute maximium and minimum of the entire graph, but there can several local maximum and minimums throughout the graph. Local maximums and minimums occur when the point f(c) is the highest or lowest point respectively in a neighborhodd of c. The first and second derivatives can be used to tell if a point is a local maximum or local minimum. Figure 7 shows a function f(x) from x=a to x=e. This graph has to maximums, one at b and one at d. B is a local maximum because it is the highest point in that piece of he graph but d is the globalmaximum because it is the highest point on the entire graph. There are also two minimums, one at a and one at c. C is a local minimum because it is the lowest point on that part of he graph, but a is the global minimum because it is the lowest point on the entire graph. If the first derivative f’(x) goes from positive to negative at x=c then c is a local maximum, and if f‘(x) goes from negative to positive at x=c then c is a local minimum. If f’’(x) is positive, then the graph of f(x) is concave up at x=c, and if f’’(x) is negative then the graph of f(x) is concave down at x=c. If f’(c)=0 and f’’(c) is positive, then f(c) is a local minmium. If f’(c)=0 and f’’(c) is negative then f(c) is a local maximum, this is called the second derivative test.

There are several theroems of calculus, some most of which are realted to each other. For example,there are forms of the fundamental therom of calculus. The first form states that if g(x) = ∫_a^b▒〖f(x)〗 dx then ∫_a^b▒〖f(x)〗 dx=g(b)-g(a) and is a way of evaluating a definite integral. For example,

if g(x) = ∫_2^4▒x^5 dx

g(x) = ├ 1/6 x^6 ]_2^4=1/6 4^6-1/6 2^6 =672

The second form states that if g(x)= ∫_a^x▒〖f(t)〗 dt where a is a constant, then g’(x)=f(x) and is a way of evaluating an indefinte integral. This is done by plugging the variable upper limit into the equation then multiplying by the derivative of the variable upper limit. So

if g(x)= ∫_a^x▒t^2 dt then g’(x) = x2

and if g(x) = ∫_a^(x^2)▒t^2 dt then g’(x) = 2×5

Two other calculus therom that are cosely related are intermediate value therom and the mean value therom. The intermediate value therom states that if the function f(x) is continous on the closed interval [a,b] and y is a number between f(a) and f(b), then there is a number, x=c, between a and b for which f(c)=y. In figure 8, if x=0 were to represent a and x=3 were to represent b, it is clearly seen that the function is continuous on the closed interval [a,b], then f(a)=0 and f(b)=-27 so there is a value for x=c between a and b for which f(c)=y. The mean value therom takes this idea one step farter. It states that if f(x) is differentiable on the open interval (a,b) and continuous on the closed interval [a,b], then there is a at least one value for x=c in (a,b) where f’(c) =(f(b)-f(a))/(b-a). This theorem can be said more simpley that if the function is differentiable and conituous, then the rate of change at c will be equal to the rate of change between a and b.

One last important relatinship is the one between opposite integrals. The integral is the opposite of , so equals – .

For example,

∫_0^2▒〖x^2 dx〗= ├ x^3/3]_0^2 = 2^3/3- 0^3/3=8/3 and ∫_2^0▒〖x^2 dx〗= ├ x^3/3]_2^0 = 0^3/3- 2^3/3=-8/3.

All of the previously mentioned ideas can be used to solve the following two problems. 1) Let f be a function defined on the closed interval -5 ≤ x ≤ 5 with f(1) = 3. The graph of f ’, the derivative of f, consists of two semicircles and two line segments, as shown in figure.

a) For -5 < x < 5, find all values x at which f has a relative maximum. Justify with calculus.

f’(x)=0 at -3, 1, and 4, f’(x) goes from positive to negative at -3 and at 4 so f has relative maximums at x=-3 and x=4

b) For -5 < x < 5, find all values x at which the graph of f has a point of inflection. Justify.

f’(x) goes from changes directions at -4, -1, and 2 so there are points of inflection at

x=-4, x=-1, and x=2

c) Find all intervals on which the graph of f is concave up and also has positive slope. Justify.

The graph of f would be concave up when the graph of f’(x) is increasing and positive, so when -5d) Find the absolute minimum value of f(x) over the closed interval -5 ≤ x ≤ 5. Justify.

The absolute minimum could be in the spot where f’(x)=0 and where f’’(x) is negative, or when f’(x) goes from negative to positive which is x=1 so f(1) =3, Or it could be at one of the endpoints, can’t forget those guys.

e) Let g be the function given by g(x) = Find g(3), g’(3), and g”(3). Justify.

g(3)= area under the curve from f’(1) to f’(3)=.5(1)(2)+.5(1+2)*1=2.5 square units

g’(3)=f’(3) = 1

g’’(3)=f’’(3)= the rate of change at f’(3) =-1

2) The functions F and G are differentiable for all real numbers, and G is strictly increasing. The table below gives values of the functions and their first derivatives at selected values of x. The function H is given by H(x) = F(G(x)) – 6

X F(x) F’(x) G(x) G’(x)

1 6 4 2 5

2 9 2 3 1

3 10 -4 4 2

4 -1 3 6 7

a) Use calculus concepts to explain why there must be a value r for 1 < r < 3 such that H(r) = -5

Using the intermediate value theorm:

h(1)=f(g(1))-6=f(2)-6=9-6=3

h(3)=f(g(3))-6=f(4)-6=-1-6=-7

h(3)<-5b) Use calculus concepts to explain why there must be a value c for 1 < c < 3 such that H’(c) = -5

Using the Mean value theorem:

h’(x)=(h(b)-h(a))/(b-a)

h’(c) = (h(3)-h(1))/(3-1) = (-10)/2 = -5

h is continuous on the open interval of (1,3) and differentiable on the closed interval [1,3], it exists that for 1c) Let w be the function given by w(x) = . Find the value of w’(3)

By using the fundamental theorem of calculus for indefinite integrals,

w’(x)=f(g(x))*g’(x)

w’(3)=f(g(3)*g’(3)

w’(3)=f(4)*2

w’(3)=-1*2= -2

d) If G-1 is the inverse function of G, write an equation for the line tangent to the graph of

y = G-1(x) at x = 2.

g(1)=2 so g-1(2)=1

g-1’=1/(g'(1))=1/5

so: y=1/5(x-2)+1

Calculus has many rules, many exceptions, and many theorems that sound and look at lot alike. The trick to knowing the difference is knowing exactly what each one does and how it works. One little trick that works well for the difference in intermediate value theorem and mean value theorem: means deal with average and dividing. That’s it! Simple, but never forgotten. Remember: This is just a sample from a fellow student.

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