AP Calc BC Chapter 5
When using the second deriv test, if you get 0 or undefined when plugging a critical point into the second derivativr
Second deriv test failed- use first deriv test
Using y=2x³-14x²+22x, do the second deriv test to find extreme values
So you already found the critical points (where first deriv is 0 or undefined) in first problem. That's x=1 and x=11/3 Then plug those into the second deriv, 12x-28 For x=1 you get a negative- that means that's a maximum For 11/3 you get a positive, that's a minimum
What is concave up? Concave down?
Up looks like a smile, down like a frown
What is a point of inflection? What would you see at the point of inflection?
Where concavity changes. You would see a sign change in the second derivative
What is a critical point
Where the derivative of a function is undefined, or where it is equal to 0 So where the original line has a horizontal or vertical slope, OR a sharp turn/discontinuity
What does "find the extreme values" mean Where do these extreme values happen
Look for absolute and local max and mins Extreme values (maxes and mins) happen at critical points (so where deriv is undefined or where it's 0) or at endpoints
Find the extreme values of f(x)= 5-2x² when x≤1 f(x)= x+2 when x>1
Look for places where deriv is undefined, or where it's 0 So deriv of 5-2x² is -4x, when x≤1 Deriv of x+2 is just 1, when x>1 So derivative does not exist at x=1 (Left hand deriv is -4 at that point, right hand deriv is 1) That means x=1 is a critical point Critical points are also where slope is 0. Derivative -4x=0 when x=0 So x=0 is a critical point Now need to know if these points are maxes or mins Plug in numbers to the left and right x=0, or f(0), is a LOCAL max At x=1, or f(1), is a LOCAL min
What if they ask "are we guaranteed to have" _______
Might be mean value theorem Check if it's continuous and diff If not, then say "no, not guaranteed to have"
Do infinite discontinuites count for absolute minima/maxima?
NO If you have an infinite discontinuity upwards, then there is no absolute maxima anywhere If you have an infinite discontinuity downwards, there's no absolute minima anywhere
Find C using MVT on [-1, 1] f(x)= x³+3 when x<1 f(x)=x²+1 when x≥1
Not continuous at x=1 1³+3=4 1+1=2 Left limit ≠ right limit MVT does not apply
What do you need to remember for cone related rates problems
Ratio of radius to height of a cone is always constant
What is a relative maxima? A relative minima?
Relative max- Any part of the graph that's higher than the values on both sides Relative min- any part of the graph that's lower than the values on both sides
What do you do if they ask you to find all points of discontinuity of the derivative from a graph of the original function
Sketch the derivative Be aware of places where the derivative goes from having a very pos slope, to 0, to pos again- though it may not look like it, that's probably a discontinuity
x(t)=t²-4t+3 x(t) describes the position, find v and a and describe the particle's motion when t>0
So find v, it's 2t-4 Then do first deriv test, you find that 2 is critical point, 1st deriv is negative until 2 and then positive after Since the velocity is negative before 2 and then becomes positive, you can say particle slows down before t=2 and then speeds up after
Find point C using the MVT for [-1, 1] √x² +1
So that's just |x|+1 Not diff at x=0 (graph it) Not diff at a place on the interval, so MVT does not apply
Water runs into a conical tank at a rate of 9 ft³/min. The tank stands vertex down and has a height of 10 feet and a base radius of 5 feet. How fast is the water level rising when the water is 6 feet deep
So the first thing you need to know is that r=1/2 h And that's a constant And the equation for volume is V=1/3πr²h So based on the information you're given, if you just took the derivative you'd have two variables to solve for So before taking the deriv, recognize that r=1/2 h is a constant, and plug that into V=1/3πr²h Simplify to v= (π/12)h³ and take deriv from there
What is a critical point that's not a stationary point
Stationary point is when the derivative is equal to 0 So "critical point that's not a stationary point" is just when the function is undefined for any reason (vertical tangent, jump/step, corner, etc) She actually said she doesnt care about stationary points
What does the mean value theorem state
That between point a and b, if the function is always continuous and always diff (REMEMBER TO WRITE THAT YOU CHECKED FOR THAT), then there's at least one point c, where the deriv or instantaneous slope is equal to the overall slope between a and b
Can endpoints be relative mins and maxes?
Yes but I think she said don't worry about that
If you have one absolute min, and one absolute max, and you don't have any other points that are relative mins/maxes, what are your relative mins/maxes?
Your absolute max would also be your relative max!!! absolute min would also be your relative min
If given f(x) and asked for F(x)
Take the antiderivative
If there were 2 points that were the lowest, would you have 2 absolute mins
yes
Show that the function f(x)=x² satisfies the hypothesis of the Mean Value Theorem on the interval [0, 2]. Then find a solution C
1. Check if continuous and diff 2. Find secant line slope, f(2)-f(0) / 2-0 = 2 3. Need deriv of x² 4. Set deriv = to SLOPE of secant 2x=2 x=1 So x=1 is the point where slope of deriv= slope of secant=overall slope between A and B
How do you find the logistic regression of some data and graph it, and superimpose the regression curve
1. Enter data in L1, L2 from stat, edit 2. Turn stat plot on 3. Stat, calc, logistic regression Then to graph it, go zoom,stat
How do you solve related rates problems
1. Make a sketch 2. Label your rates (if something is a rate, that's a derivative), your variables, and your constants 3. Find an equation- if it's a more complicated geometry one, it'll be given 4. Put in your constants (things that WONT CHANGE) 5. Do implicit diff with respect to time, so dsomething/dt 6. Plug in your variables- you should know everything except one. If you have multiple unknowns, try to solve for some of those unknowns (like maybe an unknown can be found through pythagorean theorem) 7. Solve for unknowns AND LABEL UNITS
How do you solve max/min real life problems
1. Write down what you know and what you're solving for 2. Determine the variable you're going to use for the thing that needs to be maximized or minimized, and write an equation MAY NEED PRIMARY AND SECONDARY EQUATION that just means z=2x+3 and y=2z+3x Plug in z, y=2(2x+3)+3x 3. Find the domain of the function (in other words, are there any places that dont exist that therefore CANT be maxes or mins 4. Identify critical points, and solve like you've been doing
How do you do the second derivative test
1. find critical points (where first deriv is 0 or undefined) 2. find secrond derivative 3. Plug those critical points into the second deriv. If the second deriv is positive, the function is concave up and looks like a smile, you have a min If the second deriv is negative, concave down looks like a frown, you have a max
Where is f(x)= x³-4x increasing and decreasing
1. find deriv, it's 3x²-4 2. Find critical points, they're ±√(4/3) 3. Set up intervals around critical points, and find deriv in intervals to see if increasing/decreasing Increasing on (-∞, -2/√3) (2√3, ∞) Decreasing on (-2/√3, 2/√3)
How does the intervals of concavity test work
1. find the second derivative 2. Find where second deriv is 0 or is undefined. These are POSSIBLE points of inflection 3. Put intervals around those PPOI, and plug in test values to f¹¹x to see if the sign changes If the sign does change, that's a point of inflection Points of inflection are NOT maxes or mins- they're their own thing However, this is essentially the process of the first deriv test, but with the second deriv
How are maximums and minimums described for horizontal lines
All points on a horizontal line are maximum points, and all points on a horizontal line are minimum points That means if they ask about relative mins/maxes, INCLUDE ALL HORIZONTAL LINES FOR BOTH
f(x)= √(1-x²) A=(-1, f(-1)) B=(1, F(1)) find a tangent in interval (-1, 1) that is parallel to secant AB
Check- its diff and continuous 2. Find secant, it's 0. That means this is Rohls theorem, which is when slope of secant is 0, and deriv will be 0 at a point 3. Find deriv, it's -x / √(1-x²) 4. Set deriv equal to secant x=0 5. They said find a tangent- NEED EQ find y in original eq, it's 1 Overall eq is y=1
Using y=2x³-14x²+22x, find points of inflection
Concavity test. Second deriv is y=12x-28 Find where 0 or undefined x=7/3 is possible point of inflection (-∞, 7/3) is negative second deriv so concave down (7/3, ∞) is pos, so concave up Sign changed, so 7/3 is point of inflection
Identify extreme values using analytic methods y=x^(2/5) when -3≤x≤1
Deriv is 2/ (5x^(3/5)) Never equal to 0 Undefined when 0=5x^(3/5) when x=0 Find values at -3, 0, 1 (REMEMBER ENDPOINTS) for -3, it's ⁵√(9) for 0 it's 0 for 1 it's 1 Then interval (-3, 0), derivative negative Then interval (0, 1) derivative positive So absolute max ⁵√(9) Absolute min
A rectangle is to be inscribed under one arch of the sine curve. What is the largest area the rectangle can have, and what dimensions give that area?
Draw a picture, shift the sin function π/2 to the left Call between 0 and π/2 "x" (so rectangle width is 2x) and then the shifted function is y=sin(x+(π/2)) OR cos(x) Area=2x (sin (x+(π/2)) Now solve for max Derivative is (2x)(cos(x+(π/2))+(sin (x+(π/2))(2) Never undefined- solve for when equal to 0 factor out 2 0=2(xcosx(x+(π/2))+(sin (x+(π/2))) 0=xcosx(x+(π/2))+(sin (x+(π/2)) xcosx(x+(π/2)) =- (sin (x+(π/2)) Divide out cosx(x+(π/2)) to get x alone x=- (sin (x+(π/2)) / cosx(x+(π/2)) x= -tan(x+(π/2)) use calc to solve for x, putting x in y1 and -tan(x+(π/2)) in y2 x=0.860 NOT DONE need dimensions. width is 2x, so 2(.860)=1.72 Get height with y=sin (x+(π/2) y=0.652 So max=xy=(1.72)(0.652)=1.12
Show the mean value theorem graphically
Draw any points a and b on a graph, then draw the straight line connecting them (secant line) Then find the point between a and b, where the tangent of that line is parallel to that secant
Interval is [-3,1] find all critical points for f(x)=4x⁴+10x³-5x²-2x-1 (allowed to use calc)
Find deriv, find where undefined or 0 To find zeros use calculator Then potential critical points are zeros and endpoints- find the y-values for all of those by plugging into original function for -3, y=14 for -2.13 y=33.73 for -.143 y= -.344 for .408 y=-1.85 for 1, y=6 Then make ranges between each of the points, and plug a number into the DERIVATIVE for each range to see if function increases or decreases Then determine relative mins and maxes, and absolute At x=-2.13 absolute min of -33.73 at x=-3 absolute max of 14
Find all extreme values for y= 1/ (x²-1)
Find deriv, it's -2x/ (x²-1)² Find undefined and zeros- x=0, 1, -1 Even though original function is undefined at 1 and -1, still use it to create intervals From (-∞, -1) deriv + (-1, 0) deriv + (0, 1) deriv - (1, ∞) deriv - only one extreme value- RELATIVE max at x=0 relative and not absolute bc graph has asymptotes
How do you find the point C whos deriv is parallel to the secant line (mean value theorem)
First, confirm that function is continuous on [a,b] and diff on (a,b) and write that down Need to find slope of secant, find derivative, set them equal Then solve for C
What do you do if it says "justify your answer" on a max/min problem
Include a sentence- "there is a maximum at _______ because the derivative changes from positive to negative at that point"
How can you tell that something is a related rates problem
It asks about something growing, shrinking, increasing, decreasing, changing
Find the velocity and position functions of a body falling from 0 meters accel=-9.8 m/s², falls from rest
This is taking the antiderivative, and accel is third deriv, so need first and second v(t)= -9.8t + C Solve for C if initial velocity is 0 at t=0 C=0 So just v(t)=-9.8 Then s(t)=-4.9t² +C solve for C C=0 s(t)=-4.9t²
Stadium is 518 ft high. Height after t seconds given by h=v₀t-16t². What is V₀ in order for the ball to reach a max height of 518 feet? How long will it take to reach that height?
This is a position function- take derivative to find velocity h¹=v₀-32t Need a critical point to find max velocity (or the time where velocity is max 0=v₀-32t t=v₀/32 Now plug that back into the ORIGINAL function and solve for v₀ Once you have v₀, just solve for t using t=v₀/32
Identify extreme values using analytic methods y=e^(-x) at -1≤x≤1
analytic=no graphing Deriv is -e^(-x) or -1/(e^x) Where is deriv 0 or undefined? Never 0, and denom is never 0 so no critical points BUT NEED TO CHECK ENDPOINTS- those can be extreme values at x=-1, the ORIGINAL FUNCTION value is e^(1), must be absolute max at x=1, the function y-value is e^(-1) or 1/e, must be absolute min
Find two numbers whose sum is 20 and whose product is as large as possible
x+y=20----> get in terms of y=20-x m=xy Plug in y=20-x x(20-x) m=20x-x² m¹=20-2x Find max- deriv never undefined, zero when x=10 Then do first derivative test Less than 10, deriv + Right of 10, deriv - So it is a maximum, BUT NOT DONE- solve for y, because they want 2 numbers x+y=20---->y=10 So the two numbers are 10 and 10
Find critical points of y=2x³-14x²+22x using first deriv test
y prime= 6x²-28x+22 Defined everywhere, zeros at 11/3 and 1 Find y coordinates Write out intervals Derivative increasing neg infinity to 1 decreasing 1 to 11/3 Increasing 11/3 to infinity relative max is 5 at x=1 Relative min at 11/3
