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.
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
Check my arithmetic
for example
a cos 2 pi = 4pi - pi = pi
should be
4 pi - pi = 3 pi
A.) To ensure that g(x) is a differentiable function at x = 2π, the two functions defined for x ≤ 2π and x > 2π need to match at this point. This means that their values and derivatives must be equal.
At x = 2π, the first function is:
a sin(2π) + b = a(0) + b = b
The second function is:
(2π)^2 - π(2π) + 2 = 4π^2 - 2π^2 + 2 = 2π^2 + 2
So we need to solve the following system of equations to find the values of a and b:
b = 2π^2 + 2
b = a sin(2π) + b = a(0) + b = b
Since the second equation is always true, any value of a and b that satisfy the first equation will make g(x) a differentiable function. Therefore, there are infinitely many possible values for a and b.
B.) To find the equation of the tangent line to g(x) at x = 2π, we need to find the slope of the tangent line and a point on it.
The slope of the tangent line is equal to the derivative of g(x) at x = 2π.
Let's calculate the derivatives of the two functions separately:
For the first function:
g'(x) = a cos x
For the second function:
g'(x) = d/dx (x^2 - πx + 2) = 2x - π
At x = 2π, the first function's derivative is:
g'(2π) = a cos(2π) = a(1) = a
The second function's derivative is:
g'(2π) = 2(2π) - π = 4π - π = 3π
So the slope of the tangent line is equal to a = 3π.
To find a point on the tangent line, we need to find the value of g(2π):
g(2π) = 2π^2 + 2
Therefore, the equation of the tangent line to g(x) at x = 2π is:
y - (2π^2 + 2) = (3π)(x - 2π)
C.) We can use the equation of the tangent line found in Part B to approximate the value of g(6).
Plugging in x = 6 into the equation:
y - (2π^2 + 2) = (3π)(6 - 2π)
This gives us an approximation for g(6).
A.) To find the values of a and b such that g(x) is a differentiable function, we need to ensure that the two pieces of the function match up at x = 2π.
First, we'll find the value of g(x) for x = 2π using both pieces of the function:
For the first piece: g(2π) = a sin(2π) + b = a(0) + b = b
For the second piece: g(2π) = (2π)^2 - π(2π) + 2 = 4π^2 - 2π^2 + 2 = 2π^2 + 2
Since g(x) is a continuous function, the two pieces must give the same value at x = 2π. Therefore, we have the equation: b = 2π^2 + 2.
Next, we need to find the derivative of g(x) separately for the two pieces of the function:
For the first piece: g'(x) = a cos(x)
For the second piece: g'(x) = 2x - π
To ensure differentiability, the derivatives must also match up at x = 2π. Therefore, we have the equation: a cos(2π) = 2(2π) - π.
Simplifying cos(2π) gives us a = 6π - π = 5π.
Therefore, the values of a and b that make g(x) a differentiable function are a = 5π and b = 2π^2 + 2.
B.) To find the equation of the tangent line to g(x) at x = 2π, we need to find the slope of the tangent line first.
The slope of the tangent line is given by the derivative of g(x) evaluated at x = 2π. We already found the derivatives for the two pieces of the function:
For the first piece: g'(x) = a cos(x)
For the second piece: g'(x) = 2x - π
Evaluating the derivatives at x = 2π gives us:
For the first piece: g'(2π) = 5π cos(2π) = 5π(1) = 5π
For the second piece: g'(2π) = 2(2π) - π = 4π - π = 3π
The slope of the tangent line at x = 2π is the same as the derivative of g(x) at x = 2π, so it is equal to 5π.
Now, we can use the point-slope form of a line to write the equation of the tangent line. We have the point (2π, g(2π)) = (2π, 2π^2 + 2), and the slope is 5π. The equation of the tangent line is:
y - (2π^2 + 2) = 5π(x - 2π)
C.) To write an approximation for the value of g(6) using the tangent line equation from part B, we substitute x = 6 into the equation and calculate the value.
Using the equation of the tangent line:
y - (2π^2 + 2) = 5π(x - 2π)
We can simplify it to:
y = 5π(x - 2π) + 2π^2 + 2
Now substitute x = 6 into the equation:
y = 5π(6 - 2π) + 2π^2 + 2
Simplifying this expression will give you an approximation for the value of g(6), without simplifying further.