Find any intercepts, relative extrema, points of inflection, and asymptotes. (If an answer does not exist, en?
Did you mean: y =( x^2)/(x^2 + 243 ) On which part are you having trouble? intercepts: plug in zero for x ; then plug in zero for y: (x, y) = (0,0) is the only intercept relative minimum : find the first derivative: set =0 A fraction is zero when the numerator is zero. Y' =[ (x^2+ 243)(2x) -(x^2)(2x)]/(x^2+243)^2 = (486x)/(x^2+243)^2 486x= 0 X= 0 (x, y) = (0,0) relative maximum (none) (x, y) = ________ points of inflection : set the second derivative = 0 [(X^2+243)^2* 486 - (486x)* 2(x^2+ 243)(2x)] / (x^2+ 243)^4= 0 Set the numerator = 0 X= +/-9 Plug in to the original function to find y: (x, y) =(-9, 0.25) (smaller x-value) (x, y) =(9, 0.25) larger x-value) Find the equation of the asymptote_____________ There is no vertical asymptote, since the denominator cannot= 0. For the horizontal, simplify the fraction of the largest degrees terms in the numerator and denominator: (x^2)/(x^2) Y= 1 I hope this helps!
F(x) = 2x^3-3x^2-36x Find relative extreme and inflection points?
F(x) = 2x^3-3x^2-36x F'(x) = 6x^2-6x-36 = 0 x^2-x-6=0 (x-3)(x+2)=0 x=3; x=-2; are critical points F''(x) = 12x-6 At x=3, F''(x) = 12(3)-6 = 30 > 0 , F has a relative minimum Minimum = F(3) = 2(3)^3-3(3)^2-36(3) = -81 At x=-2, F''(x) = 12(-2)-6 = -30 < 0, F has a relative maximum Maximum = F(-2) = 2(-2)^3 -3(-2)^2-36(-2) = 44 Inflection points; set the second derivative equal to 0 F''(x) = 12x-6 = 0 x= 6/12=1/2 is the inflection point
How do you find the relative extrema and asymptotes?
plug that into a graphing calculator. Now, the highest point you see in your calculator screen is the "local maxima" same with the lowest visible point, called the "local minma." Now do you see the lines being straight instead of curved? Those are asymptotes. If it a horizontal line, find the Y-value that it is on, and that is the asymptote for that line. Just, "y=5." And its the same with every other asymptote. just "x= " and "y= "
Analyze and sketch a graph of the function. Find any intercepts, relative extrema, points of inflection, and a?
x-intercept means y = 0 0 = (x^2 + 9)/(x^2 - 49) 0 = (x^2 + 9) -9 = x^2 ---> no real solutions so DNE relative minimum ---> DNE y ' = [ (x^2 - 49) * 2x - (x^2 + 9) * 2x ] / [ (x^2 - 49)^2 ] 0 = [ 2x^3 - 98x - 2x^3 - 18x ] / [ (x^2 - 49)^2 ] 0 = [ 2x^3 - 98x - 2x^3 - 18x ] / [ (x^2 - 49)^2 ] ---> so far x≠ -7 & 7 0 = - 116x 0 = x ---> critical point check this critical point whether it is a max or min into the first derived equation. +++------- _______ . .. 0. . . f(0) = ( 0^2 + 9 ) / (0^2 - 49) = -9/49 relative maximum (x , y) = ( 0 , -9/49 ) point of inflection = DNE Vertical asymptotes x = -7 x= 7 Horizontal asymptote: lim (x^2 + 9)/(x^2 - 49) ---> large in charge x---> ∞ lim (x^2/x^2) x---> ∞ lim 1 = 1 x---> ∞ y = 1 <--- horizontal asymptote ======= free to e-mail if have a question :) you are welcome
First of all, let’s focus on the maxima. Note that as [math]x\to\infty[/math], [math]x^4\to\infty[/math] while [math]4x^{-4}\to0[/math], so [math]x^4+4x^{-4}\to\infty[/math]. This is also true for [math]x\to-\infty[/math]. Additionally, as [math]x\to0[/math], [math]x^4\to0[/math] while [math]4x^{-4}\to\infty[/math], so [math]x^4+4x^{-4}\to\infty[/math]. Thus there are no relative or global maxima, as the functions continually increases as [math]x\to\pm\infty[/math] and [math]x\to0[/math].Minima, on the other hand, do exist. Luckily, we don’t even need calculus to find where they are. Instead, we can use the AM-GM Inequality, which says for positive integers [math]n_1, n_2, \dots {n_k}[/math]:[math]\dfrac{n_1+n_2+\dots+n_k}{k}\ge\sqrt[k]{n_1n_2\dots{n_k}}[/math].In our case, there are only two terms: [math]x^4[/math] and [math]4x^{-4}[/math]. Thus, using the AM-GM Inequality, we get that:[math]\dfrac{x^4+4x^{-4}}{2}\ge\sqrt{x^4\cdot4x^{-4}}[/math][math]\Rightarrow x^4+4x^{-4}\ge2\sqrt4=4[/math]Thus the minimum value of [math]y[/math] is [math]4[/math]. Now we just have to find for what values of [math]x[/math] this is the case.Let’s set another variable, [math]z[/math], to be equal to [math]x^4[/math]. This means the equation we are now trying to solve is [math]4=z+\dfrac{4}{z}[/math]. Rearranging a bit, we get [math]z^2-4z+4=0[/math]. This is just a quadratic, which we know we can solve! In fact, it can be factored as [math](z-2)^2=0[/math], and so [math]z=2[/math].Therefore, subbing back in for [math]x[/math], we have that [math]x^4=2[/math]. There are two real values of [math]x[/math] that satisfy this equation, [math]x=\sqrt[4]2[/math] and [math]x=-\sqrt[4]2[/math].Thus the minima of the function are at [math](\pm\sqrt[4]2, 4)[/math].
Analyze and sketch a graph of the graph of the function. Label any intercepts, relative extrema, points of inflection and as. y=2x/9-x^2?
y = f(x) =2x/(9-x^2) is and odd function , intersect at (0,0) 3 asymptoptes equations x = 3 , x=-3 and y=0 solve f '(x) = 0 ---> local extrema ...
Analyze and sketch the graph of the function. label any intercepts, relative extrema, points of inflection,?
"label any intercepts" can be done without calculus: It means find where it cuts the y axis: f(0) =(0 - 0 + 9)/(0 - 3) = -3 so the graph cuts the vertical axis at (0, -3) For the horizontal axis, solve f(x) = 0. The only way a fraction can be 0 is for the numerator to be zero, so solve x^2 - 3x + 9 = 0 No real roots, so it doesn't cut the horizontal axis. Although the question didn't ask for asymptotes, you need to know where they are to draw the graph sensibly. The denominator is 0 when x = 3, so x = 3 is a vertical asymptote. If we write the top as x(x-3) + 9, we can divide these two terms separately by x-3 and get f(x) = x + 9/(x - 3) and as x --> infinity, 9/(x-3) --> 0 , and so f(x) --> x Therefore the line y = x is also an asymptote. To find extrema, you need to solve f '(x) = 0. i.e. x(x-6) = 0 [I'm trusting your differentiation -- haven't checked it.] Therefore there are stationary points at (0, -3) and where x = 6, i.e. at (6, 9) [f(6) = (36 - 18 + 9)/(6-3) ..... =27/3 Now I can guess that (0, -3) is a maximum turning point, and (6, 9) is a minimum turning point, and there are no inflexions, we have two branches of a hyperbola, one in the top V region formed by y = x and x = 3, the other one opposite to it in the lower region, the inverted V. But let's check formally: Using f(x) = x + 9/(x-3) we get f '(x) = 1 - 9/(x-3)^2 This is equal to the expression you got, but is much easier to differentiate again: f "(x) = 18/(x-3)^3, by using the "function of a function" rule on 1 - 9(x-3)^(-2) Now f "(0) = 18/(-27) which is negative, therefore (0, -3) is a maximum turning point. f "(6) = 18/27, positive, so (6, 9) is a minimum turning point. Those are the two "relative extrema". 18/(x-3)^3 cannot be 0, therefore there is no point of inflexion
Don't stress the fact that [math]a[/math] is not given. You can do all of your calculations in terms of [math]a[/math]. When you go to plot the curve you can draw where [math]a[/math] is on your axes and use that as a reference to plot from.To find the points where the function is increasing/decreasing you need to set the derivative equal to zero and find the critical points (you will notice that the [math]x[/math] values will be independent of [math]a[/math]). Once you have those you can evaluate the derivative on each interval you find to see if the function is increasing or decreasing on it. You will also be able to find the extrema with this information (e.g. if the function goes from increasing to decreasing then you know that you've found a maximum point)You do the same procedure on the second derivative to get information on concavity and inflection points.Asymptotes are typically vertical or horizontal. Vertical asymptotes occur when denominators approach zero and horizontal asymptotes can be found by checking the behaviour of the function as [math]x[/math] approaches [math]\pm \infty[/math].All of this give you plenty of qualitative information to sketch out what the graph of the function looks like. Be sure to label the extrema and points of inflection (in terms of [math]a[/math]) on the graph.
When first learning this process it is a very good idea to relate the graphs to their equations and notice some very interesting things.I sincerely hope this idea helps you with this topic.