Determine the intervals where the function f(x)=xe^x is concave up and where it is concave down. Please show?
f(x) = xe^x f'(x) = e^x + xe^x (first derivative by chain rule) ------------- (2) f"(x) = e^x + e^x + xe^x (from (2)) => f"(x) = 2e^x + xe^x = e^x(2 + x) Concavity Theorem: If the function f is twice differentiable at x=c, then the graph of f is concave upward at (cf(c)) if f"(c)>0 and concave downward if f"(c)<0. f"(x) = 0 at x = -2 and f"(x)<0 for x<-2 and f"(x)>0 for x>-2 So, concave up for x>-2 concave down for x<-2
In which intervals is the function y = (x^2 - 16)^6 increasing and decreasing? In which intervals is the function concave up?
So lets solve it.Domain of f(x) is R,F'(x)= 6(x^2 — 16)^5 * 2x, it shares domain with F(x). So F(x) has critical points at x=4, x=—4 and x=0, which are the values where F'(x)=0.When f'(x)>0 f(x) is increasing and when f'(x)<0 f(x) is decreasing, making use of the bolzano’s theorem, you know that if in an interval a function is continues and does not cancel, it keeps the same sing everywhere.So, you can conclude that f(x) decreases from (-oo,—4) and (0,4), and it increases from (—4,0) and (4,+oo).Repeat the same with f''(x) but in that case when f''(x)<0 f(x) is concave up and when f''(x)>0 f(x) is concave down.
Is there a way to determine the concavity of a function from the first derivative?
Inorder to understand concavity or convexity, you have to look at how the slope of the function behaves at every point on the curve. If the slope increases as the point x increases in the function f(x), then we can say its convex and if the slope decreases as x increases then it is concave.Therefore we are looking at slope of the slope i.e. the second derivative. If the second derivative of the function is greater than zero at all points then function is convex and if it is less than zero then it is concave.
What does concave and convex mean?
A convex set is one where if [math]x[/math] and [math]y[/math] are in the set, then the line segment connecting [math]x[/math] and [math]y[/math] lies entirely in the set. Formally, if [math]x \in S[/math] and [math]y \in S[/math] then [math]\alpha x + (1 - \alpha) y \in S[/math] for [math]0 \leq \alpha \leq 1[/math].A convex function is one where the epigraph (the set of points above the curve) is a convex set. It is enough to know that the line segment connecting [math](x,f(x))[/math] and [math](y,f(y))[/math] lies above the curve. Formally, [math]f(\alpha x + (1 - \alpha) y) \leq \alpha f(x) + (1 - \alpha) f(y)[/math] for [math]0 \leq \alpha \leq 1[/math].A function is concave if its negation is convex.You can think of convex functions as “bowl-shaped” and concave functions as “upside-down bowl shaped.” Some authors refer to convex functions as “concave up” and concave functions as “concave down.”
If f(x)=xe^10x find the largest interval on which f(x) is concave upward. If we write the interval as (a, infi?
f(x)=xe^10x f'(x) = e^10x + x (10) e^10x f''(x)= 10e^10x + 10e^10x+10x e^10x (10) f''(x)=20e^10x+100xe^10x =0 20e^10x[1+5x]=0 1+5x=0 x=-1/5 --- point of inflection Consider the intervals (-∞,-1/5), (-1/5,∞) The curve is concave up on (-1/5,∞) concave down on (-∞,-1/5) a=-1/5
How do I draw graph of very complex functions such as x +sin x?
-First get the domain of the given function.Now differentiate the function to check where the function is increasing and decreasing in the domain.The interval in which dy/dx is positive is the increasing interval, and interval in which dy/dx is negative is the decreasing interval.Now double differentiate the function to check where is it concave up and where concave down.Let f(x) be a differentiable function on an interval I. Assume that f '(x) is also differentiable on I. Then,(i) f(x) is concave up on I iff(double derivative is positive on I.)(ii) f(x) is concave down on I iff(double derivative is negative on I.)(iff means ‘if and only if’)Now we have four possibilities: concave up and increasing , concave up decreasing , concave down and increasing, concave down and decreasing.These are the basic things to start a graph.Now y= x+sin x =>First of all domain will be all real numbers. i.e. the whole number linedy/dx = 1+cosx, since cos x is always less than or equal to 1 .Therefore this is an always increasing graph.Now you know the graph of sin xand y= xwhat we have to do for y = x + sin x is just add two graphs ,( or add values of y obtained from two graphs) for each x ,This graph is always increasing.
Is there a better explanation on curve sketching for ever function and its' first and second derivative?
Of course, using a computer you could just have it write the graph.But doing it by hand and using the 1st and 2nd derivative are useful.Let's say you plotted 10 points as a start. If you could find the zeros, that would even be better.Setting the first derivative to 0, tells us where the local max and min are. That's very helpful.And the second derivative tells us the points of concavity - the points where the curve switches from being concave up to concave down, or vice-versa.So the 1st and second derivative can help quite a but in sketching a graph.
Calculus: Let f be a differentiable function...?
Let f be a differentiable function whose graph passes through the point (2, 1/2). For all points (x,y) the slope on the graph of y = f(x) is given by dy/dx = y^3 (3-2x). a) Find d^2/dx^2. is the graph of f concave up, down, or neither at (2,1/2). b) Write an equation of the tangent line to the graph of f at the point (2,1/2) c) Find y=f(x) by solving the separable differential equation dy/dx=y^3 (3-2x) with the initial condition f(2)=1/2
How do i find what x-values are the graph on f(x) concave up and concave down?
We can re-write f(x) as follows: f(x) = (2x - 3)/x^2 = 2/x - 3/x^2 = 2x^(-1) - 3x^(-2) Taking derivatives yields: f'(x) = (2)(-1)x^(-2) - (3)(-2)x^(-3) = -2x^(-2) + 6x^(-3) f''(x) = (-2)(-2)x^(-3) + (6)(-3)x^(-4) = 4/x^3 - 18/x^4 = (4x - 18)/x^4. Then, f''(x) > 0 whenever (4x - 18)/x^4 > 0; but since x^4 > 0 for all x, this occurs 4x - 18 > 0 <=> x > 9/2. Similarly, f''(x) < 0 whenever x < 9/2. Therefore, f(x) is concave down on (-infinity, 9/2) and concave up on (9/2, infinity). I hope this helps!
Help finding concavity?
The second derivative gives the concavity information. If it's positive, it is concave up, if it's negative it is concave down. Where it is 0 is the inflection point. d^2/dx^2 (1 / [ (3x^2) + 6] ) = (2 (3 x^2-2))/(3 (x^2+2)^3) So it is 0 when 3x^2 - 2 = 0, i.e. x = sqrt(2/3) and -sqrt(2/3) are the inflection points. You can see that it is negative if x is between the two roots, therefore concave down, and positive otherwise, where it would be concave up.