Consider { a sin x + b, if x ≤ 2π
{ x^2 - πx + 2, if x > 2π
Do you mean:
y1 = g(x) = a sin x + b, if x ≤ 2π
y2 = g(x) = x^2 - πx + 2, if x > 2π
???
if so
for y1=y2 at x = 2 pi
a sin 2 pi + b = 4 pi^2-2 pi^2 = 2 pi^2
or
b = 2 pi^2
for continuous slope at x = 2 pi
dy/dx = a cos x if if x ≤ 2π
dy/dx = 2x - π, if x > 2π
a cos 2 pi = 4pi - pi = pi
a = pi = slope at x =2pi
so
tangent line at x = 2 pi
m = 2 pi
y = 2 pi^2
so
2 pi^2 = 2 pi (2 pi) + constant
constant = - 2 pi^2
y = 2 pi x - 2pi^2
go ahead, put in 6 for x
Consider { a sin x + b, if x ≤ 2π
{ x^2 - πx + 2, if x > 2π
A.) Find the values of a and b such that g(x) is a differentiable function.
B.) Write the equation of the tangent line to g(x) at x = 2π.
C.) Use the tangent line equation from Part B to write an approximation for the value of g(6). Do not simplify.
For my practice exam. Thanks.
2 answers
Check my arithmetic
for example
a cos 2 pi = 4pi - pi = pi
should be
4 pi - pi = 3 pi
for example
a cos 2 pi = 4pi - pi = pi
should be
4 pi - pi = 3 pi