Posted by Vanessa on Friday, January 4, 2013 at 11:04am.
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
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.
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