What is the reason that second derivatives when equal to zero, fail when finding an extremum?
When the second derivative is equal to zero, we don't know if the concavity will change signs or not. If the second derivative does not change from + to - or - to +, as shown in the picture below, a relative extreme can exist.Graph of the second derivative of a function that does not change signs when the second derivative is zero.However, if the sign does change, this does not yield an extreme.The graph of the function would instead look something like this:Before x=0, the second derivative is negative, after x=0 it is positive, and at x=0 it is 0.It doesn't form an extreme but a plateau.
How do you graph a function using the first and second derivatives to identify critical points and inflection points?
given a function, in order to graph it, find where it is at t=0, the local minimum and maximum points, the regions where the function is increasing and decreasing, the points of inflection, regions where its concave up or down, and what it looks like as t approaches infinity and negative infinity....that is enough to give it a good rough sketch...
Why is the derivative of a constant function zero?
It’s zero.In very simple and crude terms, the derivative of a function is how a function f(x) changes its value as you move along x, ie. it’s the “slope” of the function at point x.Since a constant function never changes, the rate of change is also zero and its slope is zero in every point (a constant function is an horizontal line everywhere)In more precise terms, the derivative f’(x) of f(x) can be seen as the limitLim [dt → 0] f’(x) = { [f(x+dt) - f(x)] / dt }.In a constant function f(x+dt) = f(x), meaning the limit is always 0 for any x, meaning the derivative f’(x) is zero everywhere.
If the second derivative of a function equals zero at a specific point, is the point necessarily an inflection point?
Typically, yes, but not always. Typically, the second derivative is positive on one side of that point and negative on the other, and in that case, there’s an inflection point. But the second derivative could have the same signs on both sides of that point, and it wouldn’t be an inflection point.The simplest example is [math]f(x)=x^4[/math] whose second derivative is [math]f’’(x)=12x^2[/math]. The second derivative is [math]0[/math] at [math]x=0[/math], but positive on both sides of [math]x=0[/math]. There’s no inflection point there.
Why does the first derivative of a function equal zero, showing the minimum of a function?
While the first derivative of a function being 0 may occur where the function is the minimum (such as for f(x) = x² + 1), it is not necessarily so for several reasons:The function may be at the maximum value (such as f(x) = 1 − x²).The function may be merely at a point of inflection (such as f(x) = x³).You refer to “the” minimum as if you are seeking the global minimum but the first derivative technique cannot distinguish between a local minimum and the global minimum (nor maximum either) (such as f(x) = 3x⁴ − 4x³ − 12x² + 5, which has a global minimum at x = 2 and a relative minimum at x = −1, and a local maximum at x = 0).There are extensions to the technique that will tell you whether you have a local maximum, a local minimum, or a point of inflection. In general you must find all of the relative minima to determine the global minimum (and likewise for maxima).The converse situation also applies. The minimum (or maximum) of a function might occur where the derivative of the function is not 0 for two reasons:The minimum may occur where the function is not differentiable (does not even have a derivative) (such as f(x) = |x|, which has a minimum as x = 0 but the function has a sharp bend there so the derivative is not only not 0—it does not exist at all).The function’s domain is bounded and the minimum occurs at an endpoint of the domain (such as f(x) = 2x − 1 restricted to the domain of [−1; 1], which has its minimum at x = −1 and a right-derivative of 2).Do not memorize formulas as rules. Such tends to show lack of understanding and will lead to mistakes in special cases, of which there are several for the concept of extrema of functions. Learn concepts (such as what I just presented) and any needed formulas will pop out.
How do you check if the second derivative of a function is positive, zero or negative without getting the exact value
Find its zeros and discontinuities, and determine if it is positive or negative between those intervals, exactly like you would any function.Alternatively, if you have a graph of the derivative, you can tell by looking at its slope, just like you can tell if a derivitive is positive or negative by looking at the original graph.Alternatively, on the original functions graph, see where the slope is concave up or down.It all depends on which information you have available; sometimes you are looking at the second derivative to properly graph the behavior of the function and sometimes you are looking at the function to determine the properties of a second derivative.
Suppose that the second derivative of the function y=f(x) is y''=(x-5)(x+6). For what x-values does the graph?
Inflection points occur when the curve changes from concave up to concave down, or vice-versa. This is determined by setting the second derivative equal to zero. Since you have given the second derivative, we can see that inflection points occur at x = 5 and x = -6.
Determine whether the function is concave up or concave down?
I have this problem, and I just need to know a little more detail. In class, we took the 2nd derivative of the function, set equal to 0, and solved for x. This was the inflection point. My new problem is, g(x)=-x^2+3x+4 If I was to find the derivative of that function, it would be -2. I don't know how this would fit into the problem. If I set 0=-2, that wouldn't make sense to me. Do I just find the intervals with the first derivative, pick a # in the interval, and then find whether they are concave down or concave up?, or is there more to it than that?
How do you find points of inflection of a derivative function?
points of inflection are the places where the second derivative of a function equals zero. So in this case f'(x)=6x^2-10, and f''(x)=12x. Now if you set f''(x)=0 it is the same as 12x=0 and dividing both sides by 12 you have that x=0. Thus your inflection point is at x=0. Now recall that f(x)=2x^3 -10x +3 is a third degree polynomial, which looks some what like and s curve, and it does indeed have only one place where the concavity changes from up to down, so having only one inflection point makes sense. Polynomials do not have vertical asymptotes nor do they have horizontal ones. You might be thinking about rational functions, which are polynomials divided by polynomials. If that is indeed what you are refering to them, to find the vercital asymptotes you set the denominator of the function equal to zero and solve for x. These are then the vertical asymptotes. For horizontal asymptotes it is more complicated I suggest you look it up in a math book, the simplst case is like this: If the highest degrees on the numerator and the denoinator are equal then the ratio of the coefficients is the horizontal asymptote. Example f(x)=(2x^3 -10x +3)/(3x^3-17), the highest degree of the umerator is 3 which is the same as the highest degree of the denominator, so the horizontal asymptote is y=2/3. Hope this helps!