Curve Sketching

Some Important Reminders Regarding Curve Sketching

(I will post solutions to both HW problems tomorrow night)

Increasing and Decreasing Functions

The derivative of a function can tell us where the function is increasing and where it is decreasing. If

a) f'(x) > 0 on an interval I, the function is increasing on I.

b) f'(x) < 0 on an interval I, the function is decreasing on I.

The intervals of increase and decrease will occur between points where f'(x) = 0 or f'(x) is undefined. However, these points are not necessarily critical numbers because we include x even if it is not in the domain of f. We simply want to find the intervals of increase and decrease around x, even if the function is not defined at that point.

The graph to the right illustrates this theorem. From A to B, the slope of the tangent lines are all negative, so the derivative, f'(x) is negative from A to B. The theorem above states that the function is decreasing from A to B. The graph shows that the values of the function are decreasing between A and B. Similarly, the function is also decreasing between C and D. From B to C however, the slopes of the tangent lines are positive. Therefore, the derivative is positive from B to C. The graph shows that the values of the function are increasing between B and C.

Concavity and Points of Inflection

A graph is called concave upward (CU) on an interval I, if the graph of the function lies above all of the tangent lines on I. A graph is called concave downward (CD) on an interval I, if the graph of the function lies below all of the tangent lines on I.

The second derivative of a function can tell us whether a function is concave upward or concave downward. If

a) f''(x) > 0 for all x in an interval I, the graph is concave upward on I.

b) f''(x) < 0 for all x in an interval I, the graph is concave downward on I.

The intervals of concavity will occur between points where f''(x) = 0 or f''(x) is undefined. We test the concavity around these points even if they are not included in the domain of f.

The graphs below illustrate the different forms of concavity. Remember that there are two ways in which a graph can be concave upward or concave downward. Graphs A and C illustrate the types of concavity when the function is increasing on the interval, while graphs B and D illustrate concavity when the function is decreasing on the interval. These 4 graphs cover every different form of concavity.

A point P on a curve is called a point of inflection if the function is continuous at that point and either

a) the function changes from CU to CD at P
b) the function changes from CD to CU at P

Points of inflection may occur at points where f''(x) = 0 or f''(x) is undefined, where x is in the domain of f. We must test the concavity around these points to determine whether they are points of inflection.

The graph to the right illustrates a curve with a point of inflection.

The 2nd Derivative Test

The 2nd Derivative Test...so elegant it makes the 1st derivative test look like the ugly step-sister.  If you got one thing from the notes today, hopefully it was the connection between concavity and the 2nd derivative.

Recall:

• 2nd Derivative > 0, concave up
• 2nd Derivative < 0, concave down
• 2nd Derivative = 0 or DNE, point of inflection

In order to use the 2nd Derivative Test to prove points are indeed extrema you must do the following in you HW:

1. graph (use your graphing calculator)
2. find derivative, set equal to zero to locate critical points
3. find 2nd derivative & plug critical points into 2nd derivative
• If positive...then it's a minimum
• If negative...then it's a maximum
• If it's zero...use 1st derivative test (or see if it's a POI)
4. find increasing/decreasing intervals (use critical points to help identify intervals)
5. find concave up/concave down intervals (use POIs to help identify intervals)

Example 1: Find any local extrema of f(x) = x⁴ − 8 x² using the Second Derivative Test.

f′(x) = 0 at x = −2, 0, and 2. Because f″(x) = 12 x² −16, you find that f″(−2) = 32 > 0, and f has a local minimum at (−2,−16); f″(2) = 32 > 0, and f has local maximum at (0,0); and f″(2) = 32 > 0, and f has a local minimum (2,−16).

Example 2: Find any local extrema of f(x) = sin x + cos x on [0,2π] using the Second Derivative Test.

f′(x) = 0 at x = π/4 and 5π/4. Because f″(x) = −sin x −cos x, you find that and f has a local maximum at . Also, . and f has a local minimum at .

The 1st Derivative Test

The 1st Derivative Test is simple way to prove that critical points are extrema (or are not extrema in some cases).  Make sure you do the following:

1. Find critical points
2. Test points on both sides (using the 1st derivative)
3. Compare the slopes (+ or -)...if it goes - to +, it's a min..if it goes + to -, it's a max.  If it doesn't change...then it's NOT an extrema!

Don't forget to include the increasing and decreasing intervals on your HW.  Use your graphing calculator to help with this.  

Curve sketching is a little trickier.  The following will be helpful:

• f(3) = 4 means graph the point (3,4)
• f'(x) = 0 means you have a max or min
• f'(x) DNE means you have a cusp, corner, or asymptote
• f'(x) >0 means increasing and f'(x)<0 means decreasing

See video for problem #11 and #36.

The Mean Value Theorem

Rolle's Thm on a slant!  Don't forget, when solving MVT problems, you need to show three things:

1. That f(x) is cont. on [a,b]
2. That f(x) is differentiable on (a,b) 
3. That there is AT LEAST one point between a and b, such that the derivative is equal to the slope between (a, f(a)) and (b, f(b)).

Hint:  The last problem asks you to show that the car MUST have been going that fast at least once...you can do that using the MVT.  

Two MVT Problems

Rolle's Theorem

Good ole' Michael Rolle...do you know that Rolle once argued AGAINST the use of calculus.  In fact, he even wrote a critique of calculus that was NOT very flattering.  Luckily, he came around and gave calculus one of its most useful theorems: Rolle's Theorem

While doing your Rolle's Thm HW (pg. 267, 3-8) remember you must show THREE things:

1. That the endpoints are EQUAL.  Plug your left and right hand endpoints into the original f(x) to show this.
2. You must show (or at least say) that f(x) is continuous and differentiable on the interval between [a,b].
3. Find the derivative,set equal to zero, and verify that that point exists and is in between a and b.  (if a and b are your endpoints)

Rolle's Theorem & Example

Check out video below for help on a couple of HW problems and a sweet quote from Michael Rolle himself (thanks Grant):

Critical Point Quiz

Your quiz over critical points will cover:

1. Identify extrema on a graph (find extrema over different intervals...like #3 and 4 from the HW).  Remember, open intervals can NOT be considered extrema and constant functions are considered BOTH max and mins.
2. Finding critical points.  You should be able to find places that the derivative equals 0 and DNE.  Check your critical points to make sure that they are actually in the domain of the original function. If your derivative is too hard to set equal to zero and solve, graph the derivative and look at its zeros.
3. Given a a function, you should be able to use your calculator and locate extrema over given intervals.

Critical Points (Day 2)

Important things to take from today:

1. You can not have a max/min on an open interval.
2. You can use your calculator to determine extrema (HW problems 8-10 and 50, 51)
3. Finding critical #s can be difficult...make sure you compare your critical #s with the domain of the original function (HW problems 19-29 odd)

(See video for HW hints...you will have to pause it at times)

Critical Points (Day 1)

Critical points occur at two places:

1. any where g'(x) = 0 (set derivative equal to zero)
2. any where g'(x) DNE (so look for things that would make denominator zero, negative under radical, etc)

If your derivative ends up being a rational expression (fraction with variables on top and bottom), you can set TOP = 0 to satisfy 1st condition and bottom equal to 0 to satisfy 2nd condition.

Remember:  Extrema MUST occur at critical points...but critical points do NOT necessary have to produce extrema!