Calculus

determine the absolute extreme values of the function f(x)=sinx-cosx+6 on the interval 0<=x<=2pi.

This is what i did:
1.) i found the derivative of the function which is f'(x)=cosx+sinx
2.) I set f'(x)=0 and got sinx/cosx=-1 which is tanx=-1 and then x=3π/4 , 7π/4

What would I do after this?

  1. 👍 0
  2. 👎 0
  3. 👁 367
  1. You would now sub those two values into the original equation. The larger answer is your maximum the smaller value is your minimum

    1. 👍 0
    2. 👎 0
  2. well, you have calculated where the extrema occur, so just plug in the values for x and calculate f(x).

    or, you can easily see the values because

    sinx-cosx = √2(sinx/√2 - cosx/√2)
    = √2(sinx cos π/4 - sin π/4 cosx)
    = √2 sin(x-π/4)

    the max and min are 6±√2

    naturally, these occur when x-π/4 is a multiple of π/2, as you correctly calculated.

    1. 👍 0
    2. 👎 0
  3. Thanks!

    1. 👍 0
    2. 👎 0

Respond to this Question

First Name

Your Response

Similar Questions

  1. Calculus

    1. Locate the absolute extrema of the function f(x)=cos(pi*x) on the closed interval [0,1/2]. 2. Determine whether Rolle's Theorem applied to the function f(x)=x^2+6x+8 on the closed interval[-4,-2]. If Rolle's Theorem can be

  2. Please check my Calculus

    1. Given f(x)=-6/x, choose the correct statement A. The graph of f is concave upward on the interval (negative infinity, 0) B. The graph of f is concave downward on the interval (negative infinity, 0) C. The graph of f is concave

  3. Trig.......

    I need to prove that the following is true. Thanks (2tanx /1-tan^x)+(1/2cos^2x-1)= (cosx+sinx)/(cosx - sinx) and thanks ........... check your typing. I tried 30º, the two sides are not equal, they differ by 1 oh , thank you Mr

  4. calculus

    Use analytic methods to find the global extreme values of the function on the interval and state where they occur. y=x(2-x)^(1/2), -2

  1. math;)

    The equation 2sinx+sqrt(3)cotx=sinx is partially solved below. 2sinx+sqrt(3)cotx=sinx sinx(2sinx+sqrt(3)cotx)=sinx(sinx) 2sin^2x+sqrt(3)cosx=sin^2x sin^2x+sqrt(3)cosx=0 Which of the following steps could be included in the

  2. college calculus 1

    Find the local and absolute extreme values of the function on the given interval. f(x)=x^3-6x^2+9x+1, [2,4]?

  3. pre calculus

    Given that sinx= 3/5 and that x terminates in Quadrant 2 , determine the values for cosx and tanx Find Cos x/2

  4. Trigonometry

    Simplify #1: cscx(sin^2x+cos^2xtanx)/sinx+cosx = cscx((1)tanx)/sinx+cosx = cscxtanx/sinx+cosx Is the correct answer cscxtanx/sinx+cosx? Simplify #2: sin2x/1+cos2X = ??? I'm stuck on this one. I don't know what I should do.

  1. Calculus

    Consider the function f(x) = 5/4x^4 + 2/3x^3 - 5x^2 - 4x + 5 a. Find any relative extrema for f(x); be sure to label each as a maximum or minimum. You do not need to find function values; just find the x-values. b. Determine the

  2. Math

    (sinx - cosx)(sinx + cosx) = 2sin^2x -1 I need some tips on trigonometric identities. Why shouldn't I just turn (sinx + cosx) into 1 and would it still have the same identity?

  3. calculus help

    Find the derivative of the function. y = integral from cosx to sinx (ln(8+3v)) dv lower limit = cosx upper limit = sinx y'(x) = ????

  4. Trig

    Verify the identity: tanx(cos2x) = sin2x - tanx Left Side = (sinx/cosx)(2cos^2 x -1) =sinx(2cos^2 x - 1)/cosx Right Side = 2sinx cosx - sinx/cosx =(2sinxcos^2 x - sinx)/cosx =sinx(2cos^2 x -1)/cosx = L.S. Q.E.D.

You can view more similar questions or ask a new question.