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

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**, Friday, January 4, 2013 at 11:37am
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

- calculus -
**:)**, Wednesday, August 12, 2015 at 3:56pm
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**, Wednesday, August 12, 2015 at 3:57pm
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

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