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So we know how to find and solve for derivatives. But what is the significance? Derivatives can allow us to do many things, including: * Find rate of change (slope) of a curve * Determine the motion of a particle * Find the maximum and minimum values of the function in an interval

* Analyze graphs and functions * Sketch curves by hand * Optimization (solving applied minimum and maximum problems

But before we get into these topics, let’s go over some basic definitions and theorems we will use. definitions extrema | max / min values in a function | local max / min | peaks and valleys within a curve | absolute max / min | the greatest and lowest value a function obtains, either on its entire domain or closed interval | critical point | an x value for which f’(x) |

theorems extreme value theorem | if the curve is continuous on a closed interval, then a max/min exist | theorem 3.2 | extrema can only occur at critical points | mean value theorem | assume f(x) is continuous on [a,b] and differentiable on (a,b). Then there exists at least one value c in (a,b) such that f’(c) = fb-f(a)b-a | * finding the rate of change of a curve solving derivatives allow us to find the slope at a particular point on a curve. let’s start with a basic problem. find the slope of the curve y=4x2 at x = 3. at the right, we can see that the slope of the curve at x = 2 is 16.

* determining the motion of a particle when given a position function, solving derivatives allows us to find the velocity and acceleration of the particle or object at any time. here’s a flow chart as you can see, differentiation of a position function once will give you velocity, and a second time will give you acceleration.

try this example: position for an object is given by s(t) = 2t2 – 6t – 4. find the velocity and acceleration of the particle at time = 6. interpret this in the context of the problem. from the work, we can see that at time = 6, the velocity of the particle is 18, and the acceleration is 4. but we must interpret this in context. how velocity and acceleration work in horizontal particle motion, is a + value means right, and a – value means left. therefore, we conclude that: speed is increasing if a > 0 & v > 0 , or a < 0 & v < 0 (same signs) speed is decreasing if a < 0 & v > 0 , or a > 0 & v < 0 (diff. signs)

in our problem, since both velocity and acceleration are positive, we can conclude that the object’s speed is increasing. * finding the maximum and minimum values of a function derivatives allow us to find minimum and maximum values of a function in an interval. by finding critical points – points where the slope is 0 – we can then determine maxes and mins from the values of the curve at these points find the critical points of f(x) = x3 – 27x – 20. find any local extrema.

looking at the data, and the sign chart, it can be concluded that : there is a local min at x = 3 , f(3) = -74 there is a local max at x = -3 , f(-3) = 34 * optimization derivatives allow us to solve applied minimum and maximum problems. these are realistic questions, which integrate calculus, as well as algebra. There are 5 steps to solving optimization problems: 1. Identify all given quantities to be determined. Make a sketch. 2. Write a primary equation for the quantity that is to be optimized. 3. Put the primary equation in terms of one independent variable. 4. Determine the domain of the primary equation in the context of the problem. 5. Use calculus to determine the desired maximum or minimum value.

Example. A piece of wire, length L is bent into the shape of a rectangle. Which dimensions produce maximum area? the length and width which will produce the largest area with a wire with length L is x= L4 and y= L4 .…...

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