How do I find the maximum, minimum, and point of inflection of the function f(x) =-2x²-6x+5?
To find the max or min do:[math]f’(x) = 0 [/math] then solve for xTo find point of inflection do[math]f”(x) = 0[/math] then solve for xwhere:[math]f’(x) [/math] is the first derivative of [math]f(x)[/math][math]f”(x) [/math] is the second derivative of [math]f(x)[/math]To calculate derivative use this formula:n*x ^(n-1)derivative of a constant [math]k = 0[/math]
Calculus Help! Please!!! Local Max/Min & Inflection Points?
f '(x) = -16 cosx sinx - 16 cos x = -16cosx(sinx + 1); when f '(x) = 0, x = pi/2 or 3pi/2. pi.2 is a local minimum at f(x) = -16, and 3pi/2 is a maximum at f(x) = 16. f ''(x) = -16(-2sin^2 x - sinx + 1) = -16(-2sinx + 1)(sinx + 1); set equal to 0 and sin x = 1/2, so x = pi/6 or 5pi/6; it can't be 3pi/2 since there's a maximum there. The inflection points are (pi/6,-2) and (5pi/6, -2).
Local max and min, concavity, and inflection points?
http://www.webassign.net/scalc/4-3-6.gif Answer using the graph above. 1.At what value(s) of x does f have a local maximum? 2.At what value(s) of x does f have a local minimum? 3.On what interval(s) is f concave upward? 4.On what interval(s) is f concave downward? 5.What are the x-coordinate(s) of the inflection point(s) of f ?
Local Min/Max.. Inflection points?
f(x) = 4x^3+15x^2-150x+4 f'(x) = 12x^2+30x-150 solve f'(x)=0 12x^2+30x-150=0 4x^2+10x-50=0 4x^2+20x-10x-50=0 4x(x+5)-10(x+5)=0 (x+5)(4x-10)=0 x=-5; x=2.5 (a) and (b) Consider the intervals (-∞, -5),(-5,2.5),(2.5, ∞) choose any one point from each of the intervals. if f'(x) < 0 , f(x) is decreasing on that interval if f'(x) > 0 , f(x) is increasing on that interval f'(x)= 12x^2+30x-150 (-∞, -5): choose x=-6; f'(-6)=102 > 0 : f(x) is increasing on (-∞, -5) (-5,2.5): choose x=2; f'(2) =-42 < 0: f(x) is decreasing on (-5,2.5) (2.5, ∞): choose x=3; f'(3) = 48 > 0: f(x) is increasing on (2.5, ∞) f'(x)=12x^2+30x-150 f''(x) = 24x+30 At x=-2.5, f''(x) = 24(2.5)+30 = 90 > 0, f has a local minimum Local minimum is f(2.5) = -214.75 At x=-5, f''(x) = 24(-5)+30 = -90 < 0, f has a local maximum Local maximum is f(-5) = 629 e) To find the inflection point, set f''(x)=0 f''(x)=24x+30 = 0 x=-30/24 = -5/4 (-5/4, f(-5/4)) is the inflection point (-1.25, 207.125) f) Test for concavity: Consider the intervals (-∞, -5/4) ,(-5/4,∞) choose any one point from each of the intervals. if f''(x) < 0 , f(x) is concave down on that interval if f''(x) > 0 , f(x) is concave up on that interval f''(x)= 24x+30 (-∞, -5/4): choose x=-2; f''(2)=-18 < 0 ; f is concave down on (-∞ ,-5/4) (-5/4, ∞): choose x=2; f''(2)= 78 >0 ; f is concave up on (-5/4, ∞)
Calculus help. concave up/dowm, inflection points, local min/max?
if you mean f(x)=2x^3+3x^2-432x then f ' (x)=6x^2+6x-432 f ' (x)=6(x^2+x-72) f ' (x)=6(x+9)(x-8) find critical points by letting f ' (x)=0 0=6(x+9)(x-8) x=-9 or x=8 x: (-∞,-9) -9 (-9,8) 8 (8,∞) f ': + 0 - + f(x): inc dec inc a.) f(x) is increasing from (-∞,-9) U (8,∞) b.) local min at x=8 since f(x) goes from decreasing to increasing local max at x=-9 since f(x) goes from increasing to decreasing c.) f ' (x)=6x^2+6x-432 f " (x)=12x+6 inflection point when f " (x)=0 0=12x+6 x=-1/2 concave up when f " (x) is positive and that is in the interval (-1/2, ∞) concave down when f " (x) is negative and that is in the interval (-∞,-1/2)
How to find minimum values and inflection points?
to find a minimum or maximum, just find the derivative of the equation, and then equal it to zero as the following: f(x) = 8 + 3x - x^3 f(x)' = 3 - 3x^2 f(x)' = 0 3 - 3x^2 = 0 3 = 3x^2 1 = x^2 x = +/- 1 for the values -1 & +1 are useful to find the minimum and the maximum if you plug them back in the original equation as the following f(-1) = 8 + 3*(-1) - (-1)^3 = 8 -3 +1 = 6 f(1) = 8 + 3*(1) - (1)^3 = 8 + 3 - 1 = 10 therefore, the minimum is 6 and the maximum is 10 to find the inflection point, we need to find the second derivatrive and equal it to zero as the following: f '' (x) = -6x f '' (x) = 0 -6x = 0 x = 0 now we plug this point in the also the original equation to find the inflection point as the following : f(0) = 8 +3*0 - 0^3 = 8+0+0 = 8 which is the inflection point you got a and b right
Find formula for a curve given max/min and inflection, calc help!?
Hey did you know that that equation is basically the equation of the gaussian (normal; bell curve) distribution? And that the inflection points of the curve are one standard deviation away from the mean, which is obviously at x=1? So if you know all that, one way to get a and b would be to plug the mean and variance = (standard deviation)^2 = (3-1)^2 or (1-(-1))^2 = 4 into the standard form of the equation for the gaussiuan distribution y = 1/sqrt(2*pi*var) * e^[-(x-u)^2/(2*var)], and then rearrange the equation to get a and b. Simple enough right? I'll leave the algebra to you. But the fact that the inflection points are one standard deviation away from the mean isn't very well known because it usually doesn't have any practical use. Or what if you didn't recognize the form of the equation? Well, you could solve the problem by taking derivatives and finding where they're 0, since y'=0 at a maximum and y''=0 at an inflection point of a differentiable function. You'll need to use the chain rule and the product rule, and of course know that the derivative of e^x is e^x... y' = e^[−(x−a)^2/b] * -2(x−a)/b y'' = e^[−(x−a)^2/b] * (-2/b) + e^[−(x−a)^2/b] * (-2(x−a)/b)^2 = ( 4(x-a)^2/b^2 - 2/b) * e^[−(x−a)^2/b] Now, you know from the problem statement that y'(1)=0, y''(-1)=0, and y''(3)=0, so just plug in and solve for a and b. You actually only need one of the y'' equations, since you're only have two variables to solve for, and they're actually the same equation anyway because of the squareds. Anyways I'll leave the algebra to you.
Difference between an inflection point and a local max/min?
They are similar ideas. The difference is that local maxima and minima are found by setting the first derivative to zero while the inflection points are where the SECOND derivative is zero. These represent very different ideas, but you can think of inflection points as being extrema of the derivative instead of the function itself. Local max/min always come when slope is zero, because they are precisely where the function stops going up/down (if you think about what a graph looks like, every max is essentially the top of a hill, and at that point the slope shifts from positive to zero then to negative). This is how you find local extrema. So for your example, we have f '(x) = 3x^2 + x - 2. We set this equal to zero: 3x^2 + x -2 = 0 This gives us x = [-1 +/- 5] / 6 (quadratic formula), so x is either -1 or 2/3. We then have to find where it's increasing/decreasing. To do this we just choose values between. Since this function is continuously differentiable, we can just try any values. So we choose, for example, -2, 0 and 1. f '(-2) = 8, so the function is increasing before x= -1 f '(0) = -2, so the function is decreasing between our two zeroes. f '(1) = 2, so the function is increasing after x=2/3. So now we can say that there's a local maximum at x=-1 (the function is increasing, then decreasing). And there's a local minimum at x=2/3 (decreasing then increasing). For inflection points, we take the next derivative to get: f "(x) = 6x+1 We set that equal to zero: 6x+1=0 --> 6x = -1 --> x = -1/6 We can tell that before -1/6, f "(x) will be negative and after it will be positive. So we know that the function is concave downward, then has an inflection point, then is concave upward. Concavity simply describes the sort of bowl-shape of graphs. I.e. f(x) = x^2 is curved upwards in a sort of bowl. If you look at x^3 instead you'll see from 0 onward it curves upward, while everything less than that curves off downward.
What is the difference between inflection points and extrema points?
From a calculus standpoint, extrema points occur where the first derivative is zero, and inflection points occur where the second derivative is zero.From an intuitive/graphical standpoint, extrema points look like hilltops (local maximum) or valleys (local minimum) — like the vertices of [math]y=-x^2[/math] and [math]y=x^2[/math], respectively — and inflection points look more like a “wiggle”, where the graph changes from flattening out to getting steeper, or from getting steeper to flattening out, like [math](0,0)[/math] on [math]y=x^3-x.[/math]
How do you identify the minimum, maximum, and inflection points given [math]f(x)[/math], [math]f'(x)[/math], or [math]f''(x)[/math]?
We can distinguish between them using the second derivative.First draw a U-shape. This represents a minimum. Because the slope [math] f'(x) [/math] is changing from negative to positive, [math] f''(x) [/math] is positive.Now draw a hump. This is a maximum. The slope is changing from positive to negative, so [math] f''(x) [/math] is negative.Finally, graph [math] x^3 [/math] around the point x = 0. This is the most common example of an inflection point. Here our curve is changing from a hump shape, where [math] f''(x) [/math] is negative, to a U-shape, where [math] f''(x) [/math] is positive. Thus at [math] x = 0, f''(x) = 0. [/math]We can summarize our findings now.- Minimum: [math] f''(x) > 0 [/math]- Maximum: [math] f''(x) < 0 [/math]- Inflection point: [math] f''(x) = 0 [/math]Note 1: Most people call the hump shape concave and the U shape convex. Some books instead call these concave down and concave up, respectively.Note 2: Inflection points include any point at which [math] f''(x) [/math] changes sign. [math] f'(x) [/math] is not necessarily 0.