calculus

consider g(x)= {a sin x + b, if x ≤ 2pi} {x^2 - pi x + 2, if x > 2pi}
A. Find the values of a a b such that g(x) is a differentiable function.
B. Write the equation of the tangent line to g(x) at x = 2pi.
C. Use the tangent line equation from part B to write an approximation for the value of g(6). Do not simplify.

  1. 👍
  2. 👎
  3. 👁
  1. since each part of g is differentiable, we need the slopes to match up at x=2π, so that dg/dx is defined everywhere.

    dg/dx =
    a cos x for x <= 2π
    2x-π for x > 2π

    so, we need

    a cos(2π) = 2(2π)-π
    a = 3π

    g(x) is thus
    3π sinx + b
    x^2 - πx + 2

    and we need g to be continuous, so needs must

    3π*0 + b = (2π)^2 - π(2π) + 2
    b = 2π^2 + 2

    g(x) =
    3π sinx + 2π^2 + 2
    x^2 - πx + 2

    g(2π) = 2π^2+2
    g'(2π) = 3π

    so we want the line through (2π,2π^2+2) with slope 3π:

    y-(2π^2+2) = 3π(x-2π)

    when x=6,

    y-(2π^2+2) = 3π(6-2π)
    y = 18π-6π^2 + 2π^2+2
    = 18π-4π^2+2 = 19.070

    In fact, g(6) = 3π sin(6) + 2π^2 + 2 = 19.106

    1. 👍
    2. 👎
  2. sayyyyyyyyyyyy that

    f(x) = a sinx + b
    h(x) = x^2 - πx + 2

    find h(2pi) and set it equal to f(2pi). also find h'(2pi) and set it equal to f'(2pi). together that should give you enough info to find a and b, and then the rest of the problem tests other calc concepts.

    1. 👍
    2. 👎
  3. Part a:
    g(x) = { asinx + b, for x ≤ 2π
    .........{ x² - πx + 2, for x > 2π

    g must be continuous such that
    lim x→2π⁻ g(x) = lim x→2π⁺ g(x)

    Lefthand limit:
    lim x→2π⁻ (asinx + b) = asin(2π) + b = 0 + b = b

    Right hand limit:
    lim x→2π⁺ (x² - πx + 2) =
    (2π)² - π(2π) + 2
    4π² - 2π² + 2
    2π² + 2

    b = 2π² + 2

    Also, for the derivative to exist at all values of x you must also check the limit of the derivative to see that the derivative approaches the same value from both sides:
    ..........{ acosx, for x < 2π
    g'(x) = { ?, for x = 2π
    ..........{ 2x - π, for x > 2π

    lim x→2π⁻ g'(x) =
    lim x→2π⁻ (acosx) = acos(2π) = a

    lim x→2π⁺ g'(x) =
    lim x→2π⁺ (2x - π) = 2(2π) - π = 3π

    a = 3π

    So,
    g(x) = { 3πsinx + 2π² + 2, for x ≤ 2π
    .........{ x² - πx + 2, for x > 2π,

    and g'(2π) = 3π

    Part b:
    g(2π) = b = 2π² + 2
    This gives you the point (2π, 2π² + 2)

    y - (2π² + 2) = 3π(x - 2π)
    y - 2π² - 2 = 2πx - 6π²
    y = 2πx - 4π² + 2

    Part c:
    g(6) ≈ g(2π) + g'(π)(6 - 2π)
    g(6) ≈ (2π² + 2) + 3π(6 - 2π)
    g(6) ≈ 2π² + 2 + 18π - 6π²
    g(6) ≈ -4π² + 18π + 2
    g(6) ≈ 19.07

    Using a calculator, I evaluated g(6) and got 19.11

    1. 👍
    2. 👎

Respond to this Question

First Name

Your Response

Similar Questions

  1. Calculus

    Breathing is cyclic and a full respiratory cycle from the beginning of inhalation to the end of exhalation takes about 5 seconds. The maximum rate of air flow into the lungs is about 0.5 L/s. This explains, in part, why the

  2. 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

  3. Math

    3. find the four angles that define the fourth root of z1=1+ sqrt3*i z = 2 * (1/2 + i * sqrt(3)/2) z = 2 * (cos(pi/3 + 2pi * k) + i * sin(pi/3 + 2pi * k)) z = 2 * (cos((pi/3) * (1 + 6k)) + i * sin((pi/3) * (1 + 6k))) z^(1/4) =

  4. calc

    i did this problem and it isn't working out, so i think i'm either making a dumb mistake or misunderstanding what it's asking. A particle moves along the x axis so that its velocity at any time t greater than or equal to 0 is

  1. CALCULUS LIMITS

    What is the following limit? lim as n goes to infinity of (pi/n) (sin(pi/n) + sin(2pi/n) + sin(3pi/n) +...+ sin(npi/n)) = I.) lim as n goes to infinity sigma (n and k=1) of pi/n sin(kpi/n) II.) Definite integral from 0 to pi of

  2. Algebra 2

    Find the values of the inverse function in radians. sin^-1(0.65) a. 0.71+2pi n and -0.71+2pi n b. 0.71+2pi n and -3.85+2pi n c. 0.86+2pi n and -0.86 +2pi n d. -0.61+2pi n and 2.54+2pi n 2. tan^-1(0.09) a.-0.09+2pi n b. no such

  3. precalculus

    Solve 7 cos(2 theta) = 7 sin^2(theta) + 3 for all solutions 0 less than or equal to theta less than 2pi. Solve 7 sin(2 w) - 4 cos(w) =0 for all solutions 0 le w lt 2pi

  4. math

    Find intervals on which the following functions are increasing and decreasing. 1. f(x) = sin 2x : x is greater than or equal to 0 and less than or equal to pi 2. y = sin x + cos x [0,2pi]

  1. math

    Find the values of theta (0,2pi) that satisfy sin theta= square root 3 over 2 Solve 2costheta sintheta-sintheta on (0, 2pi) please I really need help in knowing how to solve these questions. quite urgent. I don't understand how to

  2. math! please help!

    For a boat to float in a tidal bay, the water must be at least 2.7 meters deep. The depth of the water around the boat, d(t), in meters, where t is measured in hours since midnight, is d(t) = 5 + 4.6 sin(0.5t). (a) What is the

  3. Inverse Trig

    How do I simplify arcsin (sin 6 pi) given the interval 0 ≤ theta < 2pi

  4. Calculus

    Consider the function f(x)=sin(1/x) Find a sequence of x-values that approach 0 such that (1) sin (1/x)=0 {Hint: Use the fact that sin(pi) = sin(2pi)=sin(3pi)=...=sin(npi)=0} (2) sin (1/x)=1 {Hint: Use the fact that sin(npi)/2)=1

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