calculus
posted by Vanessa .
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.

calculus 
Steve
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
2xπ 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(x2π)
when x=6,
y(2π^2+2) = 3π(62π)
y = 18π6π^2 + 2π^2+2
= 18π4π^2+2 = 19.070
In fact, g(6) = 3π sin(6) + 2π^2 + 2 = 19.106 
calculus 
:)
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. 
calculus 
Clementine
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
Respond to this Question
Similar Questions

Calculus
Consider the function f(x)=sin(1/x) Find a sequence of xvalues 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 if … 
Calculus
Consider the function f(x)=sin(1/x) Find a sequence of xvalues 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 if … 
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 given … 
Calculus
Find the interval(s) where the function is increasing where and it is decreasing. f(x)=sin(x+(pi/2)) for 0≤x≤2pi so my derivative is f'(x)=cos(x+(pi/2))? 
Calculus
Consider the curve defined by y + cosy = x +1 for 0 =< y =< 2pi.... a. Find dy/dx in terms of y. *I got 1/(ysiny) but I feel like that's wrong. b. Write an equation for each vertical tangent to the curve. c. find d^2y/dx^2 in … 
Calculus
A particle moves along a horizontal line so that at any time t its position is given by x(t)=costt. Time is measured in seconds and x is measured in meters. a.) Find the velocity as a function t. Use your answer to determine the velocity … 
Maths
Eq of curve is y=b sin^2(pi.x/a). Find mean value for part of curve where x lies between b and a. I have gone thus far y=b[1cos(2pi x/a)/2]/2 Integral y from a to b=b/2(ba)ab/4pi[sin(2pi b/a)sin2pi) MV=b/2[ab sin(2pi b/a)]/(ba) … 
Calculus help??
I'm not sure how to solve this and help would be great! d/dx [definite integral from 0 to x of (2pi*u) du] is: a. 0 b. 1/2pi sin x c. sin(2pi x) d. cos (2pi x) e. 2pi cos (2pi x) This is the fundamental theorem, right? 
Algebra2
Find the values of the inverse function in radians. 1. sin^1(0.65)? 
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 angle …