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

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