Extrema and Derivative Theorems

Doing left and right bounds in a calculator to find min/max for the whole interval of a function

Don't expect it to find absolute min/max by putting "guess" anywhere. It loves to seek relative mins/maxes and also usually ignores endpoints. Do 2nd calc just in case *It also sometimes mistakes and is off by some decimals

Another name for absolute mins/maxes?

Extreme Values, or extrema

The Extreme Value Theorem

If <<f>> is continuous on a closed interval [a, b], then f attains a maximum value f(c) and an absolute minimum value f(d) at some number c & d in [a, b]. *Not a two way theorem. eg: f(x) = {x^2, x =/= 0} on interval [-2, 2]. f will be discontinuous at x = 0, but there will be maxes at the end points. Doesn't need to be differentiable at every point. Just continuous and, of course, on a closed interval. eg: y = |x| isn't differentiable at x = 0 but is continuous. Likewise at an open interval like y = x will cause problems due to being able to get infinitely closer to the excluded point. Extreme value theorem is about extreme values (extrema).

Definition of Limit

In general, lim x->a (f(x)) = L If we can move the values of f(x) arbitrarily close to L by taking X to the sufficiently close to a, but not equal to a *for there to be a limit, f does not have to be defined at a. For it to be continuous, it does have to be defined at a.

Things to know about polynomials for mins and maxes

In polynomials, f'(x) exists for all x

Tangent vs secant

Instantaneous rate of change vs average rate of change.

Significance of the mean value theorem

It differs average rate of change from instantaneous rate of change

Definition of the Derivative

f'(c) = lim_Δx -> 0 (f(c + Δx) - f(c))/Δx f'(c) = lim_x -> c (f(x) - f(c))/(x - c) if the limit exists

Define Absolute/Global Maximum

f(c) is an absolute maximum of f ON AN INTERVAL I (this is "i") if f(c) >= f(x) for all x in I (this is "i"). (This is on a DETERMINED interval)

Define Absolute/Global Minimum

f(c) is {an absolute minimum of f on the interval I (this is "i")} if {f(c) <= f(x) for all x in I (this is "i")}.

The Mean Value theorem and Formal definitions for increasing and decreasing can be used to say that

1.) If f'(x) > 0 for all x in (a, b) then f'(x) is increasing on [a, b] -If every tangent line has positive slopes, every secant line should have positive slopes 2.) If f'(x) < 0 for all x in (a, b), then f is decreasing on [a, b]. -Every secant and tangent line will be negative, 3.) If f'(x) = 0 for all x in (a, b) then f is constant on [a, b] (Can use Rolle's theorem) MVT- If I draw a secant line, there's a tangent line between the two (points of intersection?) with the same slope. Significance? We can tell where a function increasing and decreasing. This is important for when we can't see a graph. We can look at derivative to know when a function is increasing or decreasing.

Good number for rounding in calculus, and should you do the same in physics?

3 decimal places. Heck no. Significant figures and limited precision.

What is a critical number?

A critical number of a function f is a number c in the domain of f such that either f'(c) = 0 or that f'(c) does not exist.

Formal definition for decreasing behavior on a graph

A function f is increasing on an interval if for any 2 number numbers x_1 and x_2 in the interval, x_1 < x_2 implies f(x_1) > f(x_2)

Formal definition for increasing behavior of a graph

A function is increasing on an interval if for any 2 number numbers x_1 and x_2 in the interval, x_1 < x_2 implies f(x_1) < f(x_2) *Not just "can find a point where this is true." Has to be true on whole interval.

Define Strictly Monotonic

A function is strictly monotonic ON AN INTERVAL if it is either increasing or decreasing on the entire interval. (eg: g(x) = x^3)

Polynomials and rational function continuity

A polynomial is continuous everywhere. A rational function is continuous everywhere it is defined.

Proof that sign changes only occur at critical numbers?

A really long and convoluted recursive proof/explanation: You can plug in one number if you know the critical numbers. After all, you can say all points to the left and right (up to a critical number or endpoint) will share the increasing/decreasing behavior or rather positive/negative in their derivatives because if at any other point in the non-critical numbers the behavior changes from positive to negative, the derivative graph would have to cross zero or not exist at the point, so you'd know that there won't be any change from positive to negative or negative to positive in the increasing or decreasing behavior between critical numbers/endpoints. Because you can get the ONLY values where the derivative is zero or does not exist, you know that the function HAS to either be increasing (positive derivative) or decreasing (negative derivative) at every other point. Know that the derivative graph has to pass zero or not exist at a point for the function to decrease or increase. BUT that can only happen at critical numbers. We have those, so we know that the positive/negative (polarity?/sign) of the derivative (increase/decrease) won't change anywhere critical numbers aren't

Critical numbers and mins and maxes

Critical numbers are where the derivative is equal to zero or does not exist. They aren't necessarily the points where there are mins and maxes. e.g.: f(x) = x^3 Critical number at x = 0 because derivative at x = 0 is zero. f(0) if f(x) = x^3 is neither a minimum value nor a maximum value. At a min or max, the derivative is zero or the derivative does not exist. eg: for y= |x|, min at x = 0 but the derivative does not exist at x = 0. All maxes and mins are critical numbers, but not all critical numbers are mins or maxes.

(T/F) Your absolute max and min can only be at one x.

False. Looking back at the definition as well. It has greater than or equal to and less than or equal to. Your max and min can each appear at multiple x values. For example, as in a sin or cosine function.

What things can this be used to prove

Finish later

Fermat's Theorem

If <<f>> has a local minimum or maximum at c, and if f'(c) exists, then f'(c) = 0. Inversely, if f has a local min or max at c, then c is a critical number *Not a two way theorem. for f(x) = x^3, f'(0) = 0 but c = 0 is not a min or max of f. All the points to the right are higher and all the points to the left are lower. If we have a critical number there MAY be a min or max. We don't know. (Critical numbers are potential mins/maxes but not definite until checked) If c is not a critical number, then f(c) is not a local min or max. Basically, if there's an local extrema, and the derivative isn't nonexistant there, it has to be zero. Makes perfect sense for a critical number.

Mean Value Theorem

If f is a differentiable function on the interval (a, b) and continuous on [a, b], then there exists a number c in (a, b) such that f'(c) = (f(b) - f(a)) / (b - a) or, equivalently, f(b) - f(a) = f'(c)(b - a). There will be a point where the slope of the tangent line will be equal to the slope of the secant line at c? Where the average rate of change equals the instantaneous rate of change. If a function is differentiable on a open (closed) interval (a, b) and continuous on a closed interval [a, b] A number c between a and b exists such that at some x = c, the slope of the tangent line has the same slope as a secant line at c *Remember to check for the possibility of multiple tangent lines fitting this criteria. At some point c, the average rate of change of ? is equal to the instantaneous rate of change there

A limit theory (not squeeze)

If f(x) <= g(x) when x is near a (except possibly at a) and the limits of f and g both exist as x approaches a, then lim x->a (f(x)) <= lim x->a (g(x)). The graph of f(x) is lower than the graph of g(x)

Squeeze Theorem

If f(x) =< g(x) =< h(x) when x is near a (except possibly at a) and lim x->a (f(x)) = lim x->a (h(x)) = L then lim x->a (g(x)) = L Hypothesis (The "if" part of the statement): g(x) = h(x); lim x->a (f(x)) = lim x->a (h(x)) = L Conclusion (The "then" part of the statement): Lim x->a (g(x)) = L

Define Relative/Local Maximum

If there is AN OPEN INTERVAL containing c on which f(c) is a maximum, then f(c) is called a relative maximum of f. (We choose the interval here. It's like a piece of the already determined selection of the graph. We choose from the already set interval. Like a sub interval.) Open because for an absolute includes endpoints. Relative isn't an absolute

Define Relative/Local Minimum

If there is an open interval containing c on which f(c) is a minimum, then f(c) is called a relative minimum of f

The First Derivative Test

Let c be a critical number of a function f that is continuous on an open interval I (this is "i") containing c. If f is differentiable.... [ Rolle's Theorem and Mean Value Theorem have to apply ] (function is continuous and differentiable on an interval) ...on an open interval I (this is "i") containing c. If f is differentiable on the interval, except possibly at c, then f(c) can be classified as folllows//// (Duke left off here?) *Relative. Need to check endpoints for absolute

Concavity

Let f be differentiable on an open interval I. The graph of f is concave upward on I (this is "i") if f' is increasing on the interval and concave downward if f' is decreasing on the interval

When checking for absolute mins and maxes on an interval

Make a table so you'll remember. Find the endpoints first. f(a) and f(b) on an interval [a, b]. Get an equation for the derivative f'(x) Find the critical numbers starting with where they don't exist by setting the derivative equation equal to not existing. Find the critical numbers where f'(x) = 0 by setting the derivative equation equal to 0. Be sure that your critical numbers are in the interval before advancing. Cross out the ones that aren't in it. Plug in the critical numbers into the original equation. Compare f(x)'s to determine the absolute maximum and minimum values. If this is practice, check with a calculator so you know there's no greater min/max in between. You likely may not have one in test time though. There's a reason you should know there are no other mins/maxes on the interval: If there was a point that went lower than the absolute minimum, it would have to go back up at some point to meet back with the "greater" values. If this was the case, the derivative would have to hit 0 at some point, forming a critical number which would have showed up in the process of trying to find them. Changing this some words to fit with the maximum, this still applies.

Principle Square roots and other things about them

Remember that, when solving for x, applying a square root should result in a plus or minus if it's to undo an x^2. -a^2 = a^2 after all. A negative number can't have an even root. Also look up the transformations of a trig function to a graph like the period and amplitude

Rolle's theorem vs Mean value theorem

Rolle's theorem is a specific example of the mean value theorem. Rolle's theorem found first.

What does the derivative f'(c) tell is about f(x)?

Slope of the tangent line at (c, f(c)) and the instantaneous rate of change at x = c.

Rolle's Theorem

Suppose that f is a continuous function on an interval [a, b] and differentiable on (a, b). If f(a) = f(b), then there is a number c in (a, b) such that f'(c) = 0.

Derivative of the tangent?

The derivative, not second derivative The Derivative of THE SLOPE OF the tangent line is the second derivative Say the derivative THAT IS f' is equivalent to the slope of the tangent line OF f. Derivative is slope of the tangent line The tangent line's slope is the derivative The derivative of the tangent line is NOT the derivative of a derivative

Define differentiability

The domain of f' is {x | f'(x) exists}, so A function f is differentiable at c if f'(c) exists. It is differentiable on an open interval (a, b) if it is differentiable at every number in the interval. *Differentiability implies continuity. If a function is not continuous it's not differentiable, but just because a function isn't differentiable doesn't mean it's not continuous. f(x) = |x| is continuous but not differentiable at x = 0. "Because lim_x->0^(-) (|x|/x) = -1 and lim_x->0^(+) (|x|/x) = 1, g'(0) does not exist. (This is from taking the left and right limit on the graph of the derivative after using the definition for the derivative.)"

How can we find relative extrema?

Through sign analysis. If we know sign changes can only happen at critical numbers, we know that the rate of change between critical numbers will be the same for all numbers between the critical numbers. We can test a number c between critical points a and b to know whether or not the points in (a, b) are increasing or decreasing. If the points are increasing to the left of a critical number and decreasing on the right of the critical number, the critical number is a relative max (an interval is needed though to say this). If the points are decreasing to the left of a critical number and increasing on the right of the critical number, the critical number is a relative min (an interval is needed though to say this). If f'(x) changes from positive to negative, then f(x) has a maximum If f'(x) changes from negative to positive, then f(x) has a minimum *Always be careful with saying pos, neg, and constant. Functions vs Rate of change vs Derivative. If f'(x) does not change signs, then f(x) has no maximum nor minimum. *First Derivative Test. It helps to see examples G. Tables help

What can you say about positive-to-negative and negative-to-positive rates of change?

When you want to know the critical numbers of a function you want to know where the derivative is equal to zero or does not exist. If a function is changing from positive to negative or from negative to positive, the derivative has to cross the x-axis unless discontinuous. But if it does, then it should not exist there. Either output is zero or does not exist. If the derivative is going to change signs it will happen at a critical number *Not a two way thing: While a sign changes happen at critical numbers, at a critical number there MAY be a sign change.

Where is a function continuous

Where the limit as x approaches a of f(x) is equal to f(a). lim_(x->a) f(x) = f(a) This definition suggests 3 things must be true for f to be continuous at a: 1) f(a) is defined (a is in the domain of f. 2) lim_(x->a) f(x) exists (meaning that the left-hand side and right-hand side limits must equal eachother). 3) lim_(x->a) f(x) = f(a)

Finding Min or max on an open interval with a straight line or curve or something without a "camel hump" (or dip)?

You can't. If the interval is (0, 2), and f(x) = x^2, for every x close to, but not at, x = 2, a decimal number can be tacked on infinitely. The same happens at holes in graphs. There is an infinite number of numbers between numbers. There is no defined max number just before 2. In cases where there's a hump (or dip) or where the endpoint is the max value, the max is a solid, fixed, and findable number

The derivative of a min or max is....

does not exist or zero zero or does not exist

Slope of a secant line

m = ( f(b) - f(a) ) / (b - a) (think Δy/Δx) Or Slope of secant line = (f(x_0 + h) - f(x_0))/h ? For Derivative in this case f'(x) = lim_h -> 0 (f(x_0 + h) - f(x_0))/h

If you can't give a(n) ____, it's not a local max or min

open interval. This also means that endpoints will never be relative mins or maxes though they can be absolute.