Given a function f, of a variable x, the roots of the equation are values of x for which f(x) = 0.If the function, f, happens to be a polynomial function, and r is a root of f(x) then (x - r) is a factor of f(x).
this is the increasing function theorem, hope it helps "If F'(x) >= 0 , and all x's are and element of [a,b], Then F is increasing on [a,b]" use Mean Value Theorem (M.V.T) Let F'(x)>=0 on some interval Let x1< x2 (points from that interval) by M.V.T there is a point C which is an element of [x1,x2] such that F(x2)-F(x1) / X2- X1 = F'(C) this implies: F(x2)-F(x1) = F'(C) X [x2-x1] F'(C)>=0 [x2-x1]>0 therefore: F(x2)>=F(x1) Therefore: F is increasing on that interval.
The factor theorem states that for any polynomial function f(x), if f(a) = 0, then (x-a) is a factor of f(x). Let f(x) = x3-2x2-8x-5. If (x+1) is a factor, then f(-1) = 0. (x+1 = x - (-1)) Input x = -1 into f: (-1)3-2(-1)2-8(-1)-5 f(-1) = -1 -2 + 8 - 5 f(-1) = 0. Since f(-1) = 0, (x+1) is a factor of x3-2x2-8x-5. Q.E.D.
You cannot. The function f(x) = x2 + 1 has no real zeros. But it does have a minimum.
This depends on how you define your domain and codomain. f(n)=n/3 is one to one and onto when f is from R to R, but if we define f: X --> Y, where X = [0,3] and Y = [0,3], then f maps [0,3] to [0,1], so f is not onto in this case.
For an even function, f(-x) = f(x) for all x. For an odd function, f(-x) = -f(x) for all x.
A basic wave function is a sine or cosine function whose amplitude may have a value other than 1. The cosine function is an even function because it is symmetrical about the y-axis. That is, f(-x) = f(x) for all x. The sine function is an odd function because it is antisymmetrical about the y-axis. That is, f(-x) = -f(x) for all x.
It is difficult to tell what function you have in the question because the browser used by this site is hopelessly inadequate for mathematical notation.However,f(x) is an odd function of x if and only if f(-x) = -f(x) for all x.Common examples are f(x) = x^k where k is any odd integer, f(x) = sin(x).
The function given is (f(x) = -x^2). The second derivative of a function, denoted as (fβ'(x)), measures the concavity of the function. For the function (f(x) = -x^2), the first derivative (fβ(x)) is (-2x). Taking the derivative of (fβ(x)) gives us the second derivative (fββ(x)), which is (-2). So, (fβ'(x) = -2). This indicates that the function (f(x) = -x^2) is concave down for all (x), because the second derivative is negative.
A function that is symmetric with respect to the y-axis is an even function.A function f is an even function if f(-x) = f(x) for all x in the domain of f. that is that the right side of the equation does not change if x is replaced with -x. For example,f(x) = x^2f(-x) = (-x)^2 = x^2
f(f(x)) = f(x). Only if f is 1-1 then we have a solution f(x)=x.
A function f(x) is Even, if f(x) = f(-x) Odd, if f(x) = -f(-x)
If the function of the variable x, is f(x) then the roots are all the values of x (in the relevant domain) for which f(x) = 0.
If f(-x) = f(x) for all x then x is even. Example f(x) = cos(x). If f(-x) = -f(x) for all x then x is odd. Example f(x) = sin(x). In all other cases, f(x) is neither.
Looking at the graph of the function can give you a good idea. However, to actually prove that it is even or odd may be more complicated. Using the definition of "even" and "odd", for an even function, you have to prove that f(x) = f(-x) for all values of "x"; and for an odd function, you have to prove that f(x) = -f(-x) for all values of "x".
Yes, h(x) is simply a function h --> x, like f(x) is a function f --> x. The different letters are used to illustrate the fact that the two functions need not be the same.
It is difficult to be sure because the browser used for posting questions on this site is utter rubbish and strips out all mathematical symbols. If your question was f(x) = x + 2 then the inverse is f(x) = x - 2.