Find The Intervals On Which The Function Is Concave Up Or Concave Down

F(t)= sin^2(t) .. Find the intervals in which the function is concave up and down?

Here we go!

f(t)= sin^2(t)
f(t) = ( 1 - cos2t) / 2 ; cos2t = 1 - 2sin^2t
f ' (t) = sin2t
f '' (t) = 2cos2t

Finding critical points;
f ' (t) = sin2t = 0
2t = 0 or pi
t = 0 or pi/2

At t = 0
f '' (t) = 2*1 > 0 => f(t) is minimum or concave down

At t = pi/2
f '' (t) = 2*(-1) < 0 => f (t) is maximum or concave up

Find all intervals for which the graph of function concave down?

y = 8x^3 - 2x^4

Concavity is based off of the second derivative. So let's find the first derivative with the power rule:

y' = 24x^2 - 8x^3

And the second with the power rule:

y'' = 48x - 24x^2

Now set it equal to 0 to find points of concavity:

48x - 24x^2 = 0

Factor:

24x * (2 - x) = 0

Solving separately we get two possible values of x:

x = 0 and x = 2

Set up intervals:

(-infinity, 0), (0, 2), and (2, infinity)

Choose a test value in each:

x = -1, 1, and 3

Plug these into the second derivative:

48(-1) - 24(-1)^2 = -48 - 24 = -72
48(1) - 24(1)^2 = 48 - 24 = 24
48(3) - 24(3)^2 = 144 - 216 = -72

Intervals where the function is concave down have negative values, so this would be the first and third intervals, thus the solution is:

(-infinity, 0)U(2, infinity)

Find all intervals on which the function y=8x^3 - 2x^4 is concave downward? AP Calculus?

y = 8x^3 - 2x^4

y' = 24x^2 - 8x^3

y'' = 48x - 24x^2 = 24x(2 - x)

Check sign of second derivative on the intervals:

(-∞,0) test value x = -1 y'' < 0 concave down

(0,2) test value x = 1 y'' > 0 concave up

(2,∞) test value x = 3 y'' < 0 concave down

Find the intervals of concavity of F(x)=xe^-x?

F(x)=xe^-x
F'(x) = e^(-x) - xe^(-x)
F"(x) = -e^(-x) - [e^(-x) - xe^(-x)]
F"(x) = -2e^(-x) + xe^(-x)
F"(x) = [-2 + x]e^(-x)
F"(x) = 0 then
[-2 + x]e^(-x)=0 : Hint e^(-x) > 0
-2 + x =0
x = 2

if x>2 then F"(x) > 0 : Concave case
if x=2 then F"(x) = 0 : inflection point
if x<2 then F"(x) < 0 : Convex case

Help me, Find all intervals for which the graph of the function y = 8x^3-2x^4 is concave downward?

A curve is concave downward if the second derivative is less than 0.

First derivative: y'=24x^2-8x^3
Second derivative: y''=48x-24x^2

So we need to determine when 48x-24x^2<0?
First, factor: 24x(2-x)<0
Divide both sides by 24: x(2-x)<0
Multiply both sides by -1: x(x-2)>0
x*(x-2) will be positive when both terms are positive or both terms are negative:

Both terms are positive if x>0 and x-2>0
That is true if both x>0 and x>2
That is true if x>2

Both terms are negative if x<0 and x-2<0
That is true if x<0 and x<2
That is true if x<0

So, x*(x-2)>0 if x>2 or if x<0. That gives the solution set (-∞, 0) and (2, ∞ )

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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

Where is the function concave up/down?

Fimd the inflection points, these points will define your intervals.... f(x)= xe^(-x) ; f'(x) = e^(-x) - xe^(-x) ; f''(x) = -e^(-x) - e^(-x) + xe^(-x) = -2e^(-x) + xe^(-x) ; f''(x) = 0 = (x -2)e^(-x) ; x=2 ; choose a value less than 2 and test the second derivative is it positive or negative 0 is convienet f''(0) = -2 ; so on the interval (-infinity, 2) f(x) is concave down. test a poimt greater than 2 , 3for instance f''(3) = + so on the interval (2, infinity) f(x) is concave up.