Concepts of graphs of functions:

Concepts of graphs of functions: 1) Domain where the function has allowable inputs (this is looking to find math no-no s): Division by 0 (causes an asymptote) ex: f(x) = 1 x There is a vertical asymptote at x=0 in this example Domain is (-, 0) U (0, ) Square root of a negative (causes a limiting boundary) ex: f(x) = x + 5 x is > 5 or there will be an imaginary output Domain is (-5, ) Sin -1 (x value > 1) (causes a limiting boundary) ex: f(x) = sin 1 x The sin (x) function has outputs from -1 to 1 only f(x) = sin 1 x Domain is [-1, 1] Division by 0 in trig functions (causes an asymptote) ex: f(x) = tan x.tan x = sin x cos x issue when cos x = 0 (every odd π 2 units) f(x) = tan x Domain is Where k is an integer log x or ln x where x 0 (causes an asymptote) ex: f(x) = ln x Domain is (0, ) Note y = e x Boundary of outputs goes from 0 to

2) Intercepts - (where a graph intersects an axis) x- intercept is when y=0 and x coordinate has the value (a,0) or f(a) = 0 y- intercept is when x=0 and y coordinate has the value (0,b) or f(0) = b 3) Symmetry two types (even or odd) allows you to draw a mirror image across part of the graph EVEN ODD ( x, f(x)) (x, f(x)) (x, f(x)) ( x, f( x)) Y = x 2 y = x 3 Even symmetry f(-x) = f(x) plug in x for x and get outputs that are the same as original In practice with polynomials, all degrees of all of the terms must be even Example: f(x) = 2x 4 1 which is same as f(x) = 2x 4 1x 0 Symmetry across y-axis (1 st and 2 nd quadrant), and (3 rd and 4 th quadrant) Odd symmetry f(-x) = - f(x) plug in x for x and get outputs that are opposite to original In practice with polynomials, all degrees of all of the terms must be odd Example: f(x) = 4x 3 3x Symmetry across origin (1 st and 3 rd quadrant), and (2 nd and 4 th quadrant) If the degrees are mixed there is no symmetry Example: f(x) = 5x 3 + 2 Be careful with rational functions as f(x) g(x) is even if f(x)and g(x)are both odd = + Trigonometric functions: f(x) = cos x and f(x) = sec x are even all others are odd

4) Asymptotes - Do not cross lines or trend lines there are 2 types: Vertical Asymptote: Where you divided by 0 (see which way you tend toward by checking values on either side of where division by 0 occurs) VA format: x= value Horizontal Asymptote: One of three outcomes with rational functions a) Numerator greater degree - or b) Denominator greater degree 0 c) Numerator and Denominator have equal degrees coefficients of highest degree HA format: y = value (trend line, may be an asymptote may not be ) 5) Increasing or decreasing of a function- how does the graph trend over an interval Increasing- 2 ways you must know for the AP test that the function is increasing: 1) Definition- Given a graph of a function, this is where the function has output values that get bigger and bigger If this were a function graph on [0,7] it increases (0,1),(3,5),(5,7) Note 5 is a resting place 2) Calculus way to test: First derivative of function has positive value. If this were a function graph same answer, different reason the slopes are positive (0,1),(3,5),(5,7) 5 is a resting place If this were a derivative graph (or f (x)) then the function graph would increases (0,7) since it is always positive (above the x-axis) If this were f (x) then? where original function increases. (Physics: second derivative + means acceleration is +. But that could be moving in negative direction (decreasing) but slowing down ) That said,the function can be found but it is complicated to do so later in the course Decreasing - 2 ways you must know for the AP test that the function is decreasing: 1) Definition- Given a graph of a function, this is where the function has output values that get smaller and smaller (1,3) 2) Calculus way to test: First derivative of function has negative values If this were a function graph the slopes are negative (1,3) If this were a derivative graph (or f (x)) then the corresponding function graph would never decrease since it is always positive (above the x-axis) transferred allotted

6) Local Max or Min: (Absolute or Global Max values check critical numbers and endpoints on a closed interval) CRITICAL NUMBERS ( the c in f (c) ) are when DERIVATIVE = 0 OR DNE Local Max - 2 ways you must know for the AP test that the function has max: 1) Definition- Given a graph of a function, this is where the function has a bump If this were a function graph it has a local max AT 2 and AT 6. Its value is the output of the function there. This would be an absolute max as well. 2) Calculus way to test: f (c) = 0 or dne and f (c - )>0 while f (c + )<0 If this were a function graph same answer different reason note the slopes change positive to negative at 2 and at 6. YOU MUST PLUG THESE VALUES INTO THE ORIGINAL FUNCTION TO FIND MAX VALUE. If this were f (x) then the corresponding function graph would have a local max at 7 where it goes from positive to negative Sign Charting the above graph would look like this: f x - - - - - - - + + + + + + + + + + + + + + + + + - - - - - - - - - 0 0 0 1 4 7 Local Min - 2 ways you must know for the AP test that the function has max: 1) Definition- This is where the function has a bump the opposite way (at 4) 2) Calculus way to test: f (c) = 0 or dne and f (c - )<0 while f (c + )>0 If this were a function graph same answer different reason note the slopes change negative to positive at 4 on graph above. Note: 4 here is a bounce point thinking concavity (discussed later) If this were f (x) then the corresponding function graph would have a local min at 1 where it goes from negative to positive see sign chart or graph above Also - 4 would now be a resting point on the function graph Another note to consider is if this were a function graph on an open interval (like it looks) there would be no absolute minimum. However, if it were a function graph on a closed interval say from [.5, 7.5] the absolute minimum would be at the endpoints. That is why the closed interval theorem says to check the critical points and the endpoints to find the absolute maximum values of a function.

7) Concavity- Whether the function has values that are part of a trend that is concave up (like a cup) or concave down (like a cdown???) Concave up - 2 ways you must know for the AP test that the function is concave up: 1) Definition- the point(s) in question are part of it being concave up If this were a function graph from [0,9] it is concave up from (0,2), (4,6), (7.5,9) 2) Calculus way to test: f (c) >0 If this were a function graph same answer different reason Think back to the acceleration example from earlier from (0,1) slopes are negative but decreasing and from (1,2) slopes are positive and increasing. Both of these mean positive acceleration (or the rate of change of the rate of change is positive) likewise break it down from (4,5) and (5,6) and also (7.5,8) and (8,9) do you see it??? If this were f (x) then the function graph would be concave up if the slopes of this graph are positive (the rate of change of the rate of change is positive) or (slopes of the graph of the slopes are positive) or (when is the f (x) graph increasing). That occurs (1,3), (5,7), (8,9) If this were f (x) then the function is concave up when this graph is positive. (2,4), (6,9) Sign Charting the graph would look like this: f x - - - - - - - + + + + + + + + + - - - - - - - - - - - + + + + + + + 0 0 0 0 2 4 6 9 Concave down - 2 ways you must know for the AP test that function is concave down: 1) Definition- the point(s) are part of it being concave down (2,4),(6,7.5) 2) Calculus way to test: f (c) <0 If this were a function graph: from (2,3) slopes are + but decreasing and from (3,4) slopes are - and increasing. Both mean negative f (x) likewise break it down from (6,7) and (7,7.5) do you see it??? If this were f (x) then the function graph would be concave down if the slopes of this graph are negative That occurs (0,1), (3,5), (7,8) If this were f (x) function is concave down when this is negative. (0,2), (4,6)

8) Inflection Point(AP term) or Inflection Asymptote (not an AP term)- Where function has values that change how the curve turns from concave up to concave down or vice versa. Inflection point - 2 ways you must know for AP test that function has inflection point: 1) Definition- the point(s) in question where the graph changes concavity If this were a function graph from [0,9] inflection points occur at (1.5,1) and (4.5,1) 2) Calculus way to test: f (c) = 0 and f (c - )>0 while f (c + )<0 or vice versa If this were a function graph same answer different reason At 1.5 it changes from negative slopes that are increasing to negative slopes that are decreasing (from f (c) being negative to f (c) being +) At 4.5 it changes from positive slopes that are increasing to positive slopes that are decreasing (from f (c) being positive to f (c) being -) If this were f (x) then the function graph would have an inflection point when this graph changes from having positive to negative slopes or vice versa (the slopes of this graph of slopes change). At 3 (local min on the f (x)). We cannot determine the inf. pt. only where. Note: 6 turns out to be a bounce point and not an asymptote of inflection. The f (c) DNE but the original function was continuous. If this were f (x) then the function has a point of inflection at 2 and 4 and 8. Sign Charting the graph would look like this: f x + + + + + - - - - - - + + + + + - - - - - - + + 0 0 x 0 0 2 4 6 8 9 Inflection asymptote original function had an asymptote at a value. The second derivative DNE at that value. The sign chart changes at the value in question. In the above example the original function was continuous, so this is not an asymptote of inflection. f(x) = 1 is an example of an asymptote of inflection x If the sign chart of the second derivative stays + or - on both sides of a value in question then it is a bounce point.

Things to consider given algebraic equation and not a graphs: 1) When you sign chart a derivative ( f (x) ) or a second derivative ( f (x) ) you need to pick the points where f (c) or f (c) = 0 or DNE It is helpful to write a 0 for zeros and an x for DNE on the numberline 2) However: when a function is not continuous to begin with at a value, that value can t become a max or min later. The reason why f (c) failed to exist was because of continuity issues. For c to be where a max or min occurs, f (c) fails because of being at a corner. Graph example just to get a feel for what is being said Consider -1: there is an asymptote so f (-1) DNE but f(-1) is undefined as well. So -1 is not at a local max, nor asymptote of Inflection. But is a critical # to include in sign chart Consider 1: f (1) DNE but f(1) does and f(1) is a local min Consider 2: f(2) exists but the function is not continuous at 2. It is not a local max or min Equation ex: f(x) = x 1 x 2 f(0) = undefined (a vertical asymptote) f (x) = x+2 x3 f (0) = DNE f (2) = 0 then sign chart: f (x) - - - - - - - + + + + + - - - - - - x 0 0 2 Local max at 2 but not a local min at 0 This is since 0 was undefined in the original function The local max s value is 2/9 found by plugging 2 into f(x). It is important not to call 2 the local max it is the location, not the max value

It is on the closed interval [0,9] a) f increasing: because: f decreasing: b) local max (,?) because: local min (,?) because: c) concave upward: concave downward: because: because: d) Point of inflection: (,?) Also, from the wording above how do we know there are no asymptotes in the original function? e) if f(0) = 0 what does this mean? Graph:

Tell which question or statement gives the DIS A IM CI of DISAIMCIS Precalc Lim f f Domain- Statement at top says continuous f therefore domain: or (-, 0) U (0, ) Intercepts-? E) means the intercept is (0,0) Symmetry- none (a complicated graph that starts at 0 and has no reflections over any axis) Asymptotes- none (continuous function has no vertical asymptote, closed interval has no HA) Increase/Decrease-? A) increasing: (0,2), (4,6), (8,9) decreasing (2,4), (6,8) Max/Min -? B) Local max (bump) at 2. Local max (corner) at 6. Local mins (bumps) at 4 and 8 Concavity-? C) concave up: (3,6), (6,9) concave down: (0,3) Inflection-? D) inflection point at (3,?) no inflection asymptote at 6 (it is a bounce pt not an asymptote)

Using Desmos to verify graphs of first and second derivative The f graph: Cosine graph: y = a cos(bx + c) + d with c = 0 (no phase shift) pd=6, amplitude =2 (from 1 to -3 divided in ½), vertical shift 1 unit next part: looks like parabola y = a(x h) 2 + k vertex at (6,-2) and x-int (8,0) The f graph using desmos: The corresponding f " graph using desmos: f x + + + + + - - - - - - + + + + + - - - - - - + + 0 0 x 0 0 2 4 6 8 9 F x - - - - - - - - - - + + + + + + + + + + + + + + + + + 0 x 0 3 6 9 Recall from earlier physics problems if f were position, f velocity, f acceleration then when f and f : match an object is speeding up different it is slowing down The function graph shown below is : Speeding up: (2,3) {negative direction slopes getting bigger} (4,6) {positive direction slopes getting bigger} (8,9) {positive direction slopes getting bigger} Slowing down: (0,2) {positive direction slopes getting smaller} (3,4) {negative direction slopes getting smaller} (6,8) {negative direction slopes getting smaller} + direction speeding up or direction slowing down = + acceleration (position graph concave up) Negative direction speeding up or + direction slowing down = - acceleration (concave down)

Concavity related to the rate of change of the rate of change Considering concave down (negative acceleration): 2 Parts to consider: Increasing function f > 0 Decreasing function f < 0 Slowing down (slopes decreasing) Speeding up (slopes increasing) Slowing down in + direction f <0 Speeding up in a negative direction f <0 Considering concave up (positive acceleration): 2 Parts to consider: Decreasing function f < 0 Increasing function f > 0 Slowing down (slopes decreasing) Speeding up (slopes increasing) Slowing down in a negative direction f >0 Speeding up in a positive direction f >0