Calc

Let f(x)=sqrt(x) and g(x)=sin(x). Find f*g and determine where f*g is continuous on the interval (-4pi, 4pi)

  1. 👍 0
  2. 👎 0
  3. 👁 82
asked by Maria
  1. f*g = (√x)(sin x)

    Since f is not defined for x<0, that leaves us with [0,4pi)

    Both f and g are continuous everywhere defined, so there's no problem with their product.

    1. 👍 0
    2. 👎 0
    posted by Steve

Respond to this Question

First Name

Your Response

Similar Questions

  1. calculus

    Find complete length of curve r=a sin^3(theta/3). I have gone thus- (theta written as t) r^2= a^2 sin^6 t/3 and (dr/dt)^2=a^2 sin^4(t/3)cos^2(t/3) s=Int Sqrt[a^2 sin^6 t/3+a^2 sin^4(t/3)cos^2(t/3)]dt =a Int

    asked by ms on September 18, 2013
  2. Pre-Cal(Please check)

    1) Find all the solutions of the equation in the interval (0,2pi). sin 2x = -sqrt3/2 I know that the angles are 4pi/3 and 5pi/3 and then since it is sin I add 2npi to them. 2x = 4pi/6 + 2npi 2x = 5pi/3 + 2npi Now I know that I

    asked by Hannah on February 18, 2010
  3. Pre-Cal(Please check)

    1) Find all the solutions of the equation in the interval (0,2pi). sin 2x = -sqrt3/2 I know that the angles are 4pi/3 and 5pi/3 and then since it is sin I add 2npi to them. 2x = 4pi/6 + 2npi 2x = 5pi/3 + 2npi Now I know that I

    asked by Hannah on February 18, 2010
  4. Calculus

    Please look at my work below: Solve the initial-value problem. y'' + 4y' + 6y = 0 , y(0) = 2 , y'(0) = 4 r^2+4r+6=0, r=(16 +/- Sqrt(4^2-4(1)(6)))/2(1) r=(16 +/- Sqrt(-8)) r=8 +/- Sqrt(2)*i, alpha=8, Beta=Sqrt(2) y(0)=2,

    asked by COFFEE on July 10, 2007
  5. calc

    find the area between the x-axis and the graph of the given function over the given interval: y = sqrt(9-x^2) over [-3,3] you need to do integration from -3 to 3. First you find the anti-derivative when you find the

    asked by mikayla on April 18, 2007
  6. Precalculs

    I have no idea how to do these type of problems. -------Problem-------- Solve each equation on the interval 0 less than or equal to theta less than 2 pi 42. SQRT(3) sin theta + cos theta = 1 ---------------------- There is an

    asked by Kate on February 28, 2010
  7. Pre-Cal (Please help)

    Find all the soultions of the equation in the interval (0,2pi) sin 2x = -sqrt3 /2 sin -sqrt3 /2 is 4pi/3 and 5pi/3 in the unit circle 2x= 4pi/3 + 2npi 2x=5pi/3 + 2npi I do not know what to do at this point

    asked by Hannah on February 17, 2010
  8. Pre-Cal

    Find all the soultions of the equation in the interval (0,2pi) sin 2x = -sqrt3 /2 sin -sqrt3 /2 is 4pi/3 and 5pi/3 in the unit circle 2x= 4pi/3 + 2npi 2x=5pi/3 + 2npi I do not know what to do at this point.

    asked by Hannah on February 17, 2010
  9. Pre Cal.

    1. Use half-angle identity to find the exact value of cos165. MY ANSWER: (-1/2)sqrt(2+sqrt(3)) 2. Solve 2 sin x + sqrt(3) < 0 for 0

    asked by Danny on June 16, 2009
  10. math

    Eliminate the parameter (What does that mean?) and write a rectangular equation for (could it be [t^2 + 3][2t]?) x= t^2 + 3 y = 2t Without a calculator (how can I do that?), determine the exact value of each expression. cos(Sin^-1

    asked by Anonymous on August 3, 2007

More Similar Questions