Tuesday

September 2, 2014

September 2, 2014

Posted by **James** on Sunday, November 14, 2010 at 11:48pm.

f(x) = sin (x/2)

0 =< x =< 4pi

=< is supposed to be less than or equal to.

I can find the extrema, but the points of inflection has me stumped. The inflection point is (2pi,0) but shouldn't there be inflection points on x=0 and x=4pi as well?

- calculus -
**Reiny**, Monday, November 15, 2010 at 8:59amThe basic sine curve has points of inflection at every x-intercept of the curve,

y' = (1/2) cos x/2

y' = -(1/4) sin x/2

-(1/4)sin x/2 = 0

gives me x/2 = 0 or x/2 = π or x/2 = 2π etc

or

x = 0 or x = 2π or x = 4π etc

then if you domain is 0 ≤ x ≤ 4π

you are correct, there should have been points of inflection at

(0,0), (2π,0) and (4π,0)

**Answer this Question**

**Related Questions**

Calculus A - Find all relative extrema and points of inflection for the ...

Calculus A - Find all relative extrema and points of inflection for the ...

calculus - find the relative extrema and points of inflection, if possible, of y...

calculus - Find any relative extrema and any points of inflection if they exist ...

Math - For the function y=(x^4)-(2x^2)+1 Identify all relative extrema. Identify...

Math - For the function y = x^4-2x^2+1 Identify all relative extrema. Identify ...

cal - Sketch the graph and show all local extrema and inflection points. f(x)= 1...

Calc - For the function f(x) = 5x^6 + 6x^5 - 15x^4 a.) find where the function ...

math - Given the function f(x)=x^4−6x^2 a.) Find all values of x where ...

Calculus - Given the function: f(x) = x^2 + 1 / x^2 - 9 a)find y and x ...