CALC III Local Min/Max/Saddle Point question!?
1) First, find the critical points by setting the first partials equal to 0. f_x = 2y sin x, and f_y = 2y - 2 cos x. Note that 2y sin x = 0 ==> y = 0 or sin x = 0. (i) If y = 0, then 2 * 0 - 2 cos x = 0 ==> x = π/2, 3π/2 since -1 ≤ x ≤ 7. (ii) If sin x = 0, then x = 0, π, 2π If x = 0, then 2y - 2 cos 0 = 0 ==> y = 1 If x = π, then 2y - 2 cos π = 0 ==> y = -1 If x = 2π, then 2y - 2 cos 2π = 0 ==> y = 1 --------------------- So, we have five critical points (π/2, 0), (3π/2, 0), (0, 1), (π, -1), (2π, 1). We classify these with the second order partials. f_xx = 2y cos x f_xy = 2 sin x f_yy = 2. D = (f_xx)(f_yy) - (f_xy)^2 = 4y cos x - 4 sin^2(x). For (π/2, 0), (3π/2, 0), note that D = 0. So, these are saddle points. For (0, 1), (π, -1), (2π, 1), note that D = 4 > 0 and f_xx = 2 > 0 So, these are minima. ----------------------------------- 2) f_x = cos x + cos(x + y) f_y = cos y + cos(x + y). Setting these equal to 0 and subtracting, we get cos x = cos y. We have two cases: (i) If y = x: cos x + cos(2x) = 0 2cos^2(x) + cos x - 1 = 0 (2 cos x - 1)(cos x + 1) = 0 ==> cos x = 1/2 or -1 ==> x = π/3, 5π/3 or π. ---- (ii) If y = -x: cos x + cos 0 = 0 ==> cos x = -1. ==> x = π. --------------------------- Thus, we have 3 critical points (that I could find): (π/3, π/3), (5π/3, 5π/3), (π, π). Taking second order partials: f_xx = -sin x - sin(x + y) f_xy = -sin(x + y) f_yy = -sin y - sin(x + y). D = (sin x + sin(x + y))(sin y + sin(x + y)) - sin^2(x + y) For (π/3, π/3), we have D > 0 and f_xx < 0. So, this is a local maximum. For (5π/3, 5π/3), we have D > 0 and f_xx > 0. So, this is a local minimum. For (π, π), we have D = 0. So, the second derivative test is inconclusive. It is easy to check that there are values both bigger and less than f(π, π) = 5 in the vicinity of (π, π). So, this is a saddle point. ------------------------- Note: I'm curious to see what the two other critical points are... I hope this helps!
Where are the local max/min for the function 7sin(4pi x) between x=0 and x=2/4?
Assuming the upper limit is really x=2/4 (also known as 1/2), then: Let theta = 4pi x. The range of theta is now 0 to 2 pi, or the entire range of the sin function before it repeats. x = theta/4pi. Two ways to figure out where sin(theta) is maximum: (1) since theta goes over the range 0 to 2pi, we know that a sin wave reaches is maximum value (1) at pi/2 and minimum value (-1) at 3pi/2 in that range (2) to find the extrema of a function, compute the derivative of that function and set it equal to zero (e.g. find the roots over the range of the function). The derivative of sin(theta) is cos(theta). Where does cos(theta)=0 in the range 0 to 2 pi? At pi/2 and 3pi/2. By evaluating the function at those two points you find cos(pi/2) = 1 (max) and cos(3pi/2) = -1 min. Since you have an amplitude factor of 7, the local maximum is at theta=pi/2 (therefore x=(pi/2/4pi)=1/8) with a value of 7. The local minimum is at theta=3pi/2 (therefore x=3/8) with a value of -7.
Absolute & local max. and min. values?? How do you do this?? 10 POINTS guys!?
Plug in the x-values into the equation, and plot the points on a graph: f(-8) = 64 - (-8)^2 = 64 - 64 = 0 f(-7) = 64 - (-7)^2 = 64 - 49 = 15 and so on f(-1) = 64 - (-1)^2 = 64 - 1 = 63 For f(0), we plug in 0 into the 2nd equation f(0) = 4(0) - 3 = -3 f(1) = 4(1) - 3 = 1 and so on f(8) = 4(8) - 3 = 29 Absolute maximum does not exist. It would be at x = 0 if we could plug in x = 0 into f(x) = 64 - x^2, but since we can't, then there isn't an absolute max. Absolute min is at x = 0. It's value is y = -3.
Please help!! Confused about this maximum/minimum question for calculus?
Hi, To find the locations of local minimums or maximums, find the first derivative of this function. Set it equal to zero, factor it, and solve. This gives the x coordinate of local minimums or maximums. f(x) = 2x³ - 33x² + 168x + 5 f'(x) = 6x² - 66x + 168 6x² - 66x + 168 = 0 6(x² - 11x + 28) = 0 6(x - 7)(x - 4) = 0 x - 7 = 0 x = 7 x - 4 = 0 x = 4 To determine where the graph is increasing or decreasing, we will substitute numbers to the left, in between, and to the right of these numbers into 6(x - 7)(x - 4). If it comes out positive, the graph is increasing on that section. If it comes out negative, the graph is decreasing on that section. ---------------------- 4 ------------------------- 7 --------------------- ------ Let x = 0 . . . . . . . let x = 5 . . . . . . . . let x = 8 6(0 - 7)(0 - 4) . . . . 6(5 - 7)(5 - 4) . . . . . 6(8 - 7)(8 - 4) 6(-7)(-4) is positive . 6(-2)(1) is negative . 6(1)(4) is positive graph is increasing . graph is decreasing graph is increasing . . . . . . . . . . .max at 4 . . . . . . . . . . . / . . . . . \ . . . . . . . . . . / . . . . . . . \ . . . . . . . . . . . . . . / . . . . . . . . . / . . . . . . . . . \ . . . . . . . . . . . . ./ . . . . . . . . / . . . . . . . . . . .minimum at x = 7 To find the y values at the local maximum and minimum, substitute x = 4 and x = 7 into the original function, and solve. f(x) = 2x³ - 33x² + 168x + 5 f(4) = 2(4)³ - 33(4)² + 168(4) + 5 f(4) = 277 f(x) = 2x³ - 33x² + 168x + 5 f(7) = 2(7)³ - 33(7)² + 168(7) + 5 f(7) = 250 This function has a local minimum at x equals 7 with value 250 and a local maximum at x equals 4 with value 277. <==ANSWER Local Maximum (4, 277) and local minimum (7,250) I hope that helps!! :-)
Finding a cubic function given the local minimum & local maximum?
Given f(x) = ax^3 + bx^2 + cx + d: f(-2) = -8a + 4b - 2c + d = 3 f(1) = a + b + c + d = 0. At a maximum or minimum (for a polynomial), the derivative equals 0 (horizontal tangent). Since f '(x) = 3ax^2 + 2bx + c f '(-2) = 12a - 4b + c = 0 f '(1) = 3a + 2b + c = 0. ---------------------------- Now, we have 4 equations in 4 unknowns -8a + 4b - 2c + d = 3 a + b + c + d = 0 12a - 4b + c = 0 3a + 2b + c = 0. Subtracting the first two equations: 9a - 3b + 3c = -3 ==> 3a - b + c = -1. Subtracting this with the last equation: 3b = 1 ==> b = 1/3. Next, subtract the third and fourth equations. 9a - 6b = 0 ==> a = 2b/3 = 2/9. Now, 3a - b + c = -1 ==> 2/3 - 1/3 + c = -1 ==> c = -4/3. Finally, a + b + c + d = 0 ==> d = 7/9. Thus, the cubic is given by f(x) = 2x^3/9 + x^2/3 - 4x/3 + 7/9 = (1/9) [2x^3 + 3x^2 - 12x + 7]. I hope this helps!
Calculus > Local max/min of f(x) in a word problem?
Given that "k" is a positive constant, and for the purposes of this discussion, "a" is a non-negative constant (since it represents the quantity of a chemical). We start with: rate(y) = ky(a-y) First, note that if "y" is greater than "a", then (a-y) becomes negative, and the rate is negative. Similarly, if y is less than zero, then the rate also would be negative. Thus, for a non-negative rate, we require: 0 <= y <= a Put another way, this graph crosses the "y" axis (in this graph, the horizontal axis) at 0 and "a". To find the maximum rate, find the root(s) of the first derivative of the rate equation: rate(y) = ky(a-y) rate(y) = kya-ky^2 rate'(y) = ka-2ky To find the root (max/min point): 0 = ka-2ky 2ky = ka 2y = a y = a/2 At this point, the rate is: rate(y) = ky(a-y) rate(a/2) = k(a/2)(a-(a/2)) rate(a/2) = k(a/2)(a/2) rate(a/2) = ka^2/4 Also: rate''(y) = -2k Since second derivative is negative, the root we found is a maximum point. Thus, the graph of this function is a downward pointing parabola: with the vertex (max point) at (a/2, ka^2/4) crossing the horizontal axis at y=0 and y=a
I need help with this local max and local min problem, PLEASE!?!?!?!?!?
Minimize the energy equation by finding its derivative, setting that equal to zero, and solving for F. This will find the local minimum(s) and/or maximum(s). The derivative of E is 0.25 - 3.4/F^3. Setting this equal to 0 and solving for F gives us: F = 2.387. Plug 2 points (1 less than F and one greater than F) into the derivative equation to find out whether F is a maximum or minimum. Since E is negative when F < 2.387 and E is positive when F > 2.387 the point is a minimum. The bird can minimize its energy expenditure if it forages for 2.387 hours per day. I hope this helps, good luck!
Finding local max, min, and saddle points of the function?
First, compute the first order partial derivatives. f_x = -10x e^y, f_y = 5e^y (y^2 - x^2) + 5e^y (2y) = 5e^y (y^2 + 2y - x^2). Set them equal to 0 to solve for x and y: The first equation tells us that x = 0. ==> y^2 + 2y - 0 = 0 ==> y = 0 or -2. So, we have two critical points (0,0) and (0, -2). To classify these, take the second order partial derivatives. f_xx = -10 e^y, f_xy = -10x e^y f_yy = 5e^y (y^2 + 2y - x^2) + 5e^y (2y + 2) = 5e^y (y^2 + 4y - x^2 + 2). When (x,y) = (0,0): D = (f_xx)(f_yy) - (f_xy)^2 = -10 * 10 - 0^2 < 0 So, this is a saddle point. When (x,y) = (0,-2): D = (f_xx)(f_yy) - (f_xy)^2 = -10e^(-2) * -10e^(-2) - 0^2 >0 and f_xx < 0. So, this is a local maximum. I hope this helps!
How to solve Local max/min via Derivatives with degrees greater than 3?
Your 4th power polynomial is easy... factor out x^2 to get x^2(5x^2-3). Each factor can be zero, so you have x^2=0 and (5x^2-3)=0. These can easily be solved to give you the roots 0,0, +/- sqrt(3/5) In general, solving 4th power polynomials can be tricky but yours happens to factor easily. LATER: With something like the cubic equation that you have to solve, there are no easy solutions. There ARE general solutions to cubics and even quartics (4th order), but they involve a lot of work. If you believe that there should be some neat factors, then you can try the factor rule... which says that if P(x) is your polynomial, and if P(a) = 0, then (clearly) a is a root and (x-a) is a factor. You can then use polynomial division to divide out (x-a) and then you will have a simpler polynomial to solve. You could try sketching or graphing the derivative polynomial and see where it crosses the x-axis... that's another way to see approximately where the roots are - that might give you a clue to start you off using the approach above.
Max/min/concavity help! calculus?
3)The graph of a function f is given. s=5 http://www.webassign.net/scalcet/5-3-064alt.gif Consider the function below when answering the following questions. http://www.webassign.net/cgi-bin/symimage.cgi?expr=g%28x%29%3Dint_0%5Ex%20f%28t%29%20dt%2C (a) At what values of x do the local maximum and minimum values of g occur? (Enter solutions from smallest to largest. If there are any unused answer boxes, enter NONE in the last boxes.) x1min = x2min = x1max = x2max = (b) Where does g attain its absolute maximum value? (c) On what intervals is g concave downward? (Enter the intervals that contain smaller numbers first.) webassign keeps telling me that im getting it wrong. please help, its worth 11 points.