Use a graph of f(x) to determine the value of f(n), where n is a specific xvalueTable of Contents0000 Finding the value of f(2) from a graph of f(x)002Given the following graph, evaluate f(0) and solve for f(x) = 3 Solution To evaluate f(0) means to find the output of the function when the input is 0 To do this, find the point on the graph that has an xvalue of zero This will be the place where the graph crosses the yaxis For this function an input of 0 produces an output of 1On what interval is f decreasing?
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What is f(0) on a graph- Transcript Ex 51, 8 Find all points of discontinuity of f, where f is defined by 𝑓(𝑥)={ (𝑥/𝑥, 𝑖𝑓 𝑥≠0@&0 , 𝑖𝑓 𝑥=0)┤ Since we need to find continuity at of the function We check continuity for different values of x When x = 0 When x > 0 When x < 0 Case 1 When x = 0 f(x) is continuous at 𝑥 =0 if LHL = RHL = 𝑓(0) Since there are two different Consider the graph of the function f(x)=x^2x12 a) Find the equation of the secant line joining the points (2,6) and (4,0) I got the equation of the secant line to be y=x4 b) Use the Mean Value Theorem to determine a point c
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The output f (x) is sometimes given an additional name y by y = f (x) The example that comes to mind is the square root function on your calculator The name of the function is \sqrt {\;\;} and we usually write the function as f (x) = \sqrt {x} On my calculator I input x for example by pressing 2 then 5 Then I invoke the function by pressingFrom the graph of f(x), draw a graph of f ' (x) We can see that f starts out with a positive slope (derivative), then has a slope (derivative) of zero, then has a negative slope (derivative) This means the derivative will start out positive, approach 0, and then become negative Be Careful Label your graphs f or f ' appropriatelyGraph f (x)=0 f (x) = 0 f ( x) = 0 Rewrite the function as an equation y = 0 y = 0 Use the slopeintercept form to find the slope and yintercept Tap for more steps The slopeintercept form is y = m x b y = m x b, where m m is the slope and b b is the yintercept y = m x b y = m x b Find the values of m m and b b using the
Consider the function below f(x) = x1/x, x >B) Does f have a maximum or minimum value?0 (b) Explain the shape of the graph by computing the limit as x ?
Translations of Functions Suppose that y = f(x) is a function and c > 0;Here is how this function looks on a graph with an xextent of 10, 10 and a yextent of 10, 10 First, notice the x and yaxes They are drawn in red The function, f (x) = 1 / x, is drawn in green Also, notice the slight flaw in graphing technology which is usually seen when drawing graphs of rational functions with computers or0 f(x) = lim x ?
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0 and as x ?Use the given graph of y = f(x) to find the intervals on which f'(x) > 0, the intervals on which f'(x) < 0, and the values of x for which f'(x) = 0 Sketch a possible graph of y = f'(x) On what subinterval(s) is f'(x) > 0?Since f(0) = 1 ≥ 1 x2 1 = f(x) for all real numbers x, we say f has an absolute maximum over ( − ∞, ∞) at x = 0 The absolute maximum is f(0) = 1 It occurs at x = 0, as shown in Figure 412 (b) A function may have both an absolute maximum and an absolute minimum, just
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E) If f'(x)=0, then the x value is a point of inflection for f To illustrate these principles, consider the following problems 1) Suppose a) On what interval is f increasing?A quadratic function in the form f (x) = ax2 bxx f ( x) = a x 2 b x x is in standard form Regardless of the format, the graph of a quadratic function is a parabola The graph of y=x2−4x3 y = x 2 − 4 x 3 The graph of any quadratic equation is always a parabola Using the fact $f(x)>0$ on the interval where the graph is above the $x$axis, and $f(x)0$ for $x\in (3,2)\cup(0,2)\cup(3,\infty)$ b $f(x)
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So we have the graphs of two functions here we have the graph y equals f of X and we have the graph y is equal to G of X and what I want to do in this video is evaluate what G of f of F let me do the F of in another color F of negative five is f of negative five is and it can sometimes see a little daunting when you see these composite functions you're taking you're evaluating the function GThe solution set of the equation 'f(x) =0 f (x) = 0 ' is shown in purple It is the set of all values of x x for which f(x) f (x) equals zero That is, it is the set of x x intercepts of the graph The graph of a function f f is shown at rightThe Function which squares a number and adds on a 3, can be written as f (x) = x2 5 The same notion may also be used to show how a function affects particular values Example f (4) = 4 2 5 =21, f (10) = (10) 2 5 = 105 or alternatively f x → x2 5 The phrase "y is a function of x" means that the value of y depends upon the value of
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A above 1 Example f (x) = (2)x For a above 1 As x increases, f (x) heads to infinity As x decreases, f (x) heads to 0 it is a Strictly Increasing function (and so is "Injective") It has a Horizontal Asymptote along the xaxis (y=0) Plot the graph here (use the "a" slider) •If the graph intersects or touches the Xaxis at EXACTLY ONE POINT then a quadratic polynomial has TWO EQUAL ZEROES (ONE ZERO)D= b² 4ac = 0 •If the graph is either completely above Xaxis or completely below Xaxis axis ie it DOES NOT INTERSECT XAXIS axis at any point Then the quadratic polynomial HAS NO ZERO D= b² 4ac < 0 Example 22 The function f is defined by 𝑓(𝑥)={ (1−&𝑥, 𝑥0)┤ Draw the graph of f (x) For x < 0 , f(x) = 1 – x We find the points to be plotted when x < 0 For x = 0 , f(x) = 1 Hence, point to be plotted is (0,1) For x > 0 , f(x) = x 1 We fi
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The graph of y = f(x) c is obtained by translating the graph of y = f(x) c units upwards; Calculus AB Let f be the function that is given by f (x)= (axb)/ (x^2 c) It has the following properties 1) The graph of f is symmetrical with respect to the yaxis 2) The graph of f has a vertical asymptote at x=2 3) The graph of f passesConsider the function f ( x) = { 1 x if x ∈ ( 0, ∞) 0 else I know that this is the typical example of a discontinuous function whose graph is closed, but I can't see why it is closed I understand that I'm probably just being dumb here, but if x n → 0 , we have f ( x n) → ∞, so the graph
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Before we begin graphing, it is helpful to review the behavior of exponential growth Recall the table of values for a function of the form f (x) = b x f (x) = b x whose base is greater than one We'll use the function f (x) = 2 x f (x) = 2 x Observe how the output values in Table 1 change as the input increases by 1 1Graph f (x)=2x f (x) = 2x f ( x) = 2 x Rewrite the function as an equation y = 2x y = 2 x Use the slopeintercept form to find the slope and yintercept Tap for more steps The slopeintercept form is y = m x b y = m x b, where m m is the slope and b b is the yintercept y = m x b y = m x b Find the values of m m and b b using Given the graph of a function f, we can construct the graph of its antiderivative F provided that (a) we know a starting value of F, say F(a), and (b) we can evaluate the integral ∫b af(x)dx exactly for relevant choices of a and b For instance, if we wish to know F(3), we can compute F(3) = F(a) ∫3 af(x)dx
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The graph of an exponential function is a strictly increasing or decreasing curve that has a horizontal asymptote Let's find out what the graph of the basic exponential function y = a x y=a^x y = ax looks like (i) When a > 1, a>1, a > 1, the graph strictly increases as x HSBC244 has shown a nice graph that has derivative #f'(3)=0# Here are couple of graphs of functions that satisfy the requirements, but are not differentiable at #3# #f(x) = abs(x3)5# is shown below graph{y = abs(x3)5 14, 25, 616, 1185} #f(x) = (x3)^(2/3) 5# is shown on the next graphSee the answer Find all values of x such that f (x) > 0 and all x such that f (x) < 0 and sketch the graph of f 1) Answer f (x) > 0 if x < 2 or 0< x< 4 ;
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Evaluating a Limit Using a Table of Functional Values 1 Evaluate lim x → 0 sin x x using a table of functional values Show Solution We have calculated the values of f ( x) = ( sin x) / x for the values of x listed in (Figure) Table of Functional Values for lim x → 0 sin x x xSolutionShow Solution \ \text { Given } f\left ( x \right) = x^3 6 x^2 9x\ \ \Rightarrow f'\left ( x \right) = 3 x^2 12x 9\ \ \text { For a local maxima or a local minima, we must have }\ \ f'\left ( x \right) = 0\ \ \Rightarrow 3 x^2 12x 9 = 0\ \ \Rightarrow x^2 4x 3 = 0\The value of your function, "f ()", when the argument, "x", equals 0 So, for example, if you are graphing the distance you travel on a given trip, with x representing time from the start, and f (x) representing your distance from home, you would likely start the graph at the origin, (0,0) f (0) =0 at the start of the trip
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F ( 3) = 1 – ( 3) = 4 f (2) = 1 – ( 2) = 3 f ( 1) = 1 – ( 1) = 2 etc Also, f (x) = 1 for x = 0 Lastly, f ( x) = x 1 for, x > 0 and f (1) = 2, f (2) = 3, f (3) = 4, f (4) = 5 and so on Thus, the graph of f is as shown below Concept Some Functions and Their Graphs Report ErrorThis is asking me for all the values of (g o f)(x) = g(f(x)) for x = –3, –2, –1, 0, 1, 2, and 3So I'll just follow the points on the graphs and compute all the values (g o f)(–3) = g(f(–3)) = g(1) = –1I got this answer by looking at x = –3 on the f(x) graph, finding the corresponding yvalue of 1 on the f(x) graph, and using this answer as my new xvalue on the g(x) graphDefn f is concave down if the graph of f lies below the tangent lines to f Defn f is concave up if the graph of f lies above the tangent lines to f Concavity Test If f′′(x) > 0 for all x ∈(a,b), then f is concave upward on (a,b) If f′′(x) < 0 for all x ∈(a,b), then f is concave down on (a,b) Defn The point (x0,y0) is an inflection point if f is continuous at x0 and if the concavity changes at x0
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Free functions calculator explore function domain, range, intercepts, extreme points and asymptotes stepbystepSolutions to Graphing Using the First and Second Derivatives SOLUTION 1 The domain of f is all x values Now determine a sign chart for the first derivative, f ' f ' ( x) = 3 x2 6 x = 3 x ( x 2) = 0 for x =0 and x =2 See the adjoining sign chart for the first derivative, f ' Now determine a sign chart for the second derivativeGraph f(x) = −2x 2 3x – 3 a = −2, so the graph will open down and be thinner than f(x) = x 2 c = −3, so it will move to intercept the yaxis at (0, −3) Before making a table of values, look at the values of a and c to get a general idea of what the graph should look like
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The range of f is given by the interval (∞ , 1 Example 4 Find the domain of function f given below, graph it and find its range f( x ) = √ ( x 2 4) Solution to Example 4 The domain of function given above is found by solving the polynomial inequality x 2 4 ≥ 0 The solution set of the above inequality is given by the interval 2 , 2 which is also the domain of the above functionThe graph of y = f(x c) is obtained by translating the graph of y = f(x) c units to the left;Graph the linear function f given by f (x) = 2 x 4 Solution to Example 1 You need only two points to graph a linear function These points may be chosen as the x and y intercepts of the graph for example Determine the x intercept, set f (x) = 0 and solve for x 2 x 4 = 0 x = 2 Determine the y intercept, set x = 0 to find f (0)
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X is not equal to 0 or 1F(x) = (c) Use calculus to find the exact maximum value of f(x) f(x) = (d) Use a graph of f '' to estimate the xcoordinates of the inflection points (Round yourThen the graph of y = f(x c) is obtained by translating the graph of y = f(x) c units to the right;
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A necessary but not sufficient condition For a function f, if its second derivative f″(x) exists at x 0 and x 0 is an inflection point for f, then f″(x 0) = 0, but this condition is not sufficient for having a point of inflection, even if derivatives of any order exist In this case, one also needs the lowestorder (above the second) nonzero derivative to be of odd order (third, fifthSelect the correct choice below and, if necessary, fill in the answer box to complete your choice OA For this equation, The parabola opens upward because a > 0, resulting in a vertex that is a minimum The yintercept of the quadratic function f (x) = x² 3x 4 is (0, c), ie, the point From we get the xintercepts at The axis of symmetry
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The function f(x) = x 2/3 is defined for all x and differentiable for x ≠ 0, with the derivative f ′(x) = 2x −1/3 /3 Since f is not differentiable at x=0 and f'(x)≠0 otherwise, it is the unique critical point The graph of the function f has a cusp at this point with vertical tangent The corresponding critical value is f(0) = 0We draw the graphs of f(x) = x 2 − 2x − 3 and g(x) = 2x − 3 by making a table of values and plotting the points We can see, both from the graph and from the table of values, that the graphs intersect when x = 0 and x = 4 Solving algebraically x 2 − 2x − 3 = 2x − 3 x 2 − 4x = 0 x(x − 4) = 0 Giving the solutions x = 0 and x = 42) Sketch the graph of a function whose first and second derivatives are always negative 3)
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So, (1) Any line with positive slope through (2,1) meets the condition for example f (x) = 3(x − 2) 1 has f '(x) = 3 > 0 for all x ≠ 2 and also for x = 2 graph {y1=3 (x2) 1907, 919, 245, 31} Replace 3 with any positive number for another exampleF (x) < 0 if 2 < x< 0 or x > 4 2) Answer f (x) > 0 if x > 2 f (x) < 0 if x < 2 ;
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