Find all critical points for the function f(x)=sin^2x on the interval [0,pi]?
f(x)=sin^2x <<< derivative is not sin(2x) ... use substitution to help sin x = u f(u) = u^2 ......................... u(x) = sin x f '(u) = 2u du/dx .................du/dx = cos x .... now sub f '(x) = 2(sin x)(cos x) ....... now set = 0 and solve 2(sin x)(cos x) = 0 sin x = 0 .... and cos x = 0 .... on [0, pi] x = 0 and pi .... x = pi / 2 <<< critical values sub into f(x) f(0) = 0 f(pi/2) = 1 f(pi) = 0 critical points ... (0,0), (pi/2, 1), and (pi,0)
Find all critical values of f on the interval?
Let f be the function given by f(x)=(x^-7)*lnx (a)find all critical values of f on the interval[1,10] (b)find the maximum and the minimum of f on the interval[1,10] pls show yr steps 'cause I m somewhat unclear about critical values stuff
Find all the critical point of the function f(x)=2cosx+|x| on the interval (-π,π).?
Maximum and minimum points of this function exist in the point where derivative of the function equals to zero. f ' (x) = 0 The function f(x)=2cosx+|x| consists of two parts f 1(x) = 2cosx + x and f 2(x) = 2cosx - x Now we find derivatives of the both functions : f 1*(x) = --2 sinx + 1 --2sinx + 1= 0 --2sinx= --1 sinx=0,5 x= π/6 (30*) and x= 5π/6 (150*) f 2* = --2sinx -- 1 --2sinx -1 = 0 --2sinx= 1 sinx= --0,5 x= -- π/6 (--30*) and x= --5π/6 (--150*) Now we have to find the second derivative f 1**(x) = --2cosx f 1**(π/6) = --2cosπ/6 = --2 √3/2= --√3 f1**(5π/6) = --2cos5π/6= -- √3 --√3 < 0 (this is maximum) f 2**(x) = --2cosx f 2**(--π/6)=√3 f2**(-- 5π/6) =√3 √3 > 0 (this is minimum) maximum points (π/6 ; --√3) and ( 5π/6; -- √3 ) mimimum points (-- π/6 ;√3) and (-- 5π/6; √3 ) Answer: max. critical points are (π/6 ; --√3) and (5π/6; -- √3 ) min critical points are (-- 5π/6; √3 ) and (-- π/6 ;√3) I double-checked the answer,so it's correct. Good luck with math!
How does one find all extreme values on the interval?
Second question: differentiate -3 cosec 3x cot 3x = 0 = -3 cos 3x/sin^2 3x and you should be able to decide where cos 3x = 0. Also check whether the extremes are at pi/18 or pi/4.
If a function has no critical points, then how can I find where the function is increasing or decreasing? For example, f(x)=3x(3x^2 + 1)^(1/2).
If a function has no critical points, then it tells us two things:The derivative exists on all values in the interval.The derivative’s slope is never [math]0[/math] on the interval, as at critical points, the derivative is either [math]0[/math] or it doesn’t exist.Therefore, your function either has to be only increasing everywhere, or only decreasing.It can’t change from increasing to decreasing or vice versa as then the derivative/slope would either be [math]0[/math] or would not exist at some point.I hope this helps.
What is a critical point in calculus?
Let me just expand a little on the excellent response of Fabio García.First, derivatives in the classic sense only exist for a point in the interior of the domain of a function. So if we are searching for extrema of [math]f(x)[/math], then calculus helps to the extent that we include places where [math]f'(x)=0[/math] and that in turn is not defined the edges of the domain. But we want to define critical points to be all possible extrema. So what are the possible locations of an extremum on a closed interval?endpoints (simple example [math]f(x)=x[/math] which is steadily increasing on the whole real line. So for this function, extrema for a closed interval occur at the endpoints).classic critical points (interior points where the derivative is zero). Depending on other considerations, such a point may be a local extremum only, or nothing special. Places where the derivative does not exist. The absolute value of otherwise smooth functions can (usually does) create "corners" where the function suddenly changes directions. It's customary to add such points to the list of critical points. Having done so, we have an exhaustive list of where to look for extrema of a piece-wise smooth function. The usual functions are very smooth, but absolute value is a typical way that corners are introduced. Example [math]f(x)=\lvert(x-1)(x-2)(x-3)\rvert[/math] on the interval [0,4]
Given the first derivative of the function f'(x) how many critical values does it have on the open interval...
I’d suggest you another approach: (cosx)^2=0.5*(1+cos(2x)); if y’=0 then 1+cos(2x)=0.4*x or cos(2x)=0.4*x-1 (the x=0 is outside of our interval). Now look at the graphs of y1=cos(2x) and that of y2=0.4x-1; |cos(2x)|<=1; ‘start’ point of y2 is [0,-1], y2 is rather ‘slow-up’ and y2 must be <= than 1, ie 0.4*x-1<=1 or x<=5; thus our interval is not (0,10) any longer, it is (0,5]; 5=1.59*pi; 1.59 is the # of periods of y1 in 5, because 1 period of y1 is 2x=2*pi and x=pi. Thus there are only 3 critical points in the interval (0,10)!
What are the critical points f(x, y) = cos (x^2+y^2)?
A critical point of a function with two variables is a point where the partial derivatives of first order are equal to zero.The partial derivative of f(x,y) with respect to x is -2x sin(x^2+y^2). This will be zero if x=0 or if sin(x^2+y^2)=0.The partial derivative of f(x,y) with respect to y is -2y sin(x^2+y^2). This will be zero if y=0 or if sin(x^2+y^2)=0.One obvious critical point is (0,0).I will have to use “pi” from here on. Stupid machine (or stupid user).If sin(x^2+y^2)=0 then cos(x^2+y^2)=1 or cos(x^2+y^2)=-1. So in the interval 0 to 2pi for x^2+y^2, x^2+y^2=pi/2 or x^2+y^2=3pi/2.If x^2+y^2=pi/2, then x=(pi/2-y^2)^½ or x=-(pi/2-y^2)^½So you can find a value for x for any value of y^2 between -(pi/2)^½ and (pi/2)^½ or for any value of y between 0 and (pi/2)^½ (since y^2 cannot be negative).If x^2+y^2=3pi/2, then x=(3pi/2-y^2)^½ or x=-(3pi/2-y^2)^½Likewise you can find a value for x for any value of y between 0 and (3pi/2)^½Apparently the number of critical points is bigger than you can count.I have not looked at partial derivatives much since college, so my logic may be very faulty.