How can you determine the vertex of a cubic function?
You need to clarify this question: What do you mean by “vertex” here? That term is not typically used with cubic functions. Specifically: Any quadratic function can be written in “vertex form” [math]a(x-h)^2+k[/math]. However, not every cubic function can be rewritten as [math]a(x-h)^3+k[/math]; any cubic in that form (with [math]h\ne 0[/math]) would have both a linear term [math]3h^2x[/math] and a quadratic term [math]-3hx^2[/math], so (for example) [math]x^3+x^2[/math] and [math]x^3-5x[/math] cannot be written like that.Do you mean the minima/maxima? (There might be one of each, or neither.) In that case, the cubic [math]f(x)=ax^3+bx^2+cx+d[/math] would have extrema at the zeros of [math]f'(x)=3ax^2+2bx+c[/math]; i.e., at [math]x=\displaystyle\frac{-2b\pm\sqrt{4b^2-12ac}}{6a}=\frac{-b\pm\sqrt{b^2-3ac}}{3a}[/math]. If these values are complex, or there is only one zero (that is, if [math]b^2-3ac\le 0[/math]), then [math]f(x)[/math] is strictly monotone (increasing or decreasing) and has no extrema.Or do you mean the center of rotational symmetry of the cubic — which is the same as the inflection point? (Every cubic has this, and this would be somewhat closer to the idea of “vertex form” discussed earlier.) This occurs at the zero of [math]f''(x)=6ax+2b[/math], which is [math]x=-\frac{b}{3a}[/math]. The [math]y[/math]-coordinate of this point is [math]\frac{2 b^3}{27 a^2}-\frac{b c}{3 a}+d[/math].
Calclus question help please! "Use the derivative to determine whether the function is strictly monotonic
f(x) =( x^4/4) - 2x^2 df/dx = x^3 - 4x = x(x^2 - 4) = 0 The solutions of this are: x = 0. x = 2 and x = -2 There are four regions then that need to be examined to see how the derivative behaves in each of them: -Inf to -2 -2 to 0 0 to 2 2 to +Inf A monotonic function can have the derivative equal 0 at points and that is OK. For an increasing function the derivative in each of these intervals should be positive. For a decreasing function the derivative in each of these interals should be negative. Test each of these regions using a selected value of x: -Inf to -2: Use x = -10 and get df/dx < 0 So the function is decreasing over this range. -2 to 0: Use x = -1 and get df/dx > 0 So the function is increasing over this range A monotonic function is one whose successive values are either increasing or decreasing. Therefore this function is not strictly monotonic since it both increases and decreases.
Show that a function from a finite set S to itself is one-to-one if and only if it s onto?
Assume that there exists a finite set S such f:S->S is a one-to-one function but not onto. Recall the definition of onto: Given a function g:A->B, for all b that are elements of B, there exists an aEA such that g(a)=b. We are assuming our function (call it f) is not onto, thus there exists a b that is an element of S, such that there does not exist any aES such that f(a)=b. Thus, we have just shown that the image of f is at most one element smaller than that of the domain (remember, b isn't in the image). But if this is true, then because f is finite and a function, all elements in S have to map to one and only one element in S/{b}. However, we could exhaust all unique pairs, ie. f(a1)=b1, f(a2)=b2..., f(an)=bn, until we eventually arrive at a situation where one element in the domain still needs to be paired and the only element left in S (the codomain) is b. But f(a) doesn't equal b as we have shown for any aES, thus this is a contradiction. This is not necessarily true for infinite sets because the norm of an infinite set can still equal the norm of that same set minus 1 or some other finite (and in some cases infinite) number of elements.
Derivative question please help!?
forwhonow i really understand your explanations and i have a calc test tomarrow :( mind if i ask another question? what are the rules behind this one Find the derivative of the function below. y = 3tan(x^4) + sec(π/3). (dy)/(dx) = ? i assumed i can take the derivative of each smaller term and add them togethor so i can approach the problem and focus on 3tan(x^4) and sec(pi/3) or am i wrong in thinking that? also i really dont know how to go much farther than there... lol
Find the derivative of the curve y = x^3 - 3x^2 + 27x -3?
Using the sum rule, d/dx(x^3 - 3x^2 + 27x -3) distributes. Then you just use the power rule , which is d/dx y^n=n*y^(n-1), to get: d/dx(x^3 - 3x^2 + 27x -3)=3x^2-6x+27. In quadratics, b^2 - 4ac is the discriminant; this tells you the nature of the roots. Since your first derivative is quadratic opening upwards (since a>0), the function is monotonic increasing if there are no real roots, or if there are two equal roots (since it is not monotonic if the first derivative changes sign, and sign changes happen at the roots). Therefore, for it to be monotonic increasing, b^2 - 4ac must be negative or zero. Subbing in, b^2 - 4ac<=0 (-6)^2 - 4(3)(27)<=0 36-324<=0 -288<=0. Since the discriminant is negative 288, there are no real roots. Therefore, the first derivative does not change sign. Therefore, the function is monotonic increasing.
Urgent! (critical point) The function f(x) = cos x - sin x has one critical point in the region 0 <= x <= pi.?
Critical points in a function occur wherever the first derivative is zero, or is undefined (such as division by zero). f(x) = cos x - sin x f'(x) = -sin x - cos x = 0 -sin x = cos x - tan x = 1 tan x = -1 x = arctan(-1) x = -π/4 + Nπ, where N is any integer The only solution that lies within your given boundary is N=1: x = 3π/4 = 2.3562 (b) Evaluate f'(x) on both sides of the critical point. Let's use the values π/2 and π: f'(π/2) = -sin (π/2) - cos (π/2) = -(1) - 0 = -1 f'(π) = -sin (π) - cos (π) = 0 - (-1) = 1 Since f'(x) changes from a negative value to a positive one, we know the critical point is a local minimum. (c) f'(x) = -sin x - cos x f''(x) = -cos x + sin x f''(3π/4) = - cos (3π/4) + sin (3π/4) = -(-1) + 0 = 1 Because the second derivative is positive at the critical point, the function is concave upward at that point.