Answer:
Well, if <em>y </em>is 1
Then the equation would be y = 1 (2x)
So, depending on the answer options. Y would be double of whatever the X value is.
See how 1 X value is worth 2 Y?
Example Attached
If f(x)=-3x-1, find f(-5)a)16b)15c)-16d)14
1 answer:
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Take the Laplace transform of both sides:
L[y'' - 4y' + 8y] = L[δ(t - 1)]
I'll denote the Laplace transform of y = y(t) by Y = Y(s). Solve for Y :
(s²Y - s y(0) - y'(0)) - 4 (sY - y(0)) + 8Y = exp(-s) L[δ(t)]
s²Y - 4sY + 8Y = exp(-s)
(s² - 4s + 8) Y = exp(-s)
Y = exp(-s) / (s² - 4s + 8)
and complete the square in the denominator,
Y = exp(-s) / ((s - 2)^2 + 4)
Recall that
L⁻¹[F(s - c)] = exp(ct) f(t)
In order to apply this property, we multiply Y by exp(2)/exp(2), so that
Y = exp(-2) • exp(-s) exp(2) / ((s - 2)² + 4)
Y = exp(-2) • exp(-s + 2) / ((s - 2)² + 4)
Y = exp(-2) • exp(-(s - 2)) / ((s - 2)² + 4)
Then taking the inverse transform, we have
L⁻¹[Y] = exp(-2) L⁻¹[exp(-(s - 2)) / ((s - 2)² + 4)]
L⁻¹[Y] = exp(-2) exp(2t) L⁻¹[exp(-s) / (s² + 4)]
L⁻¹[Y] = exp(2t - 2) L⁻¹[exp(-s) / (s² + 4)]
Next, we recall another property,
L⁻¹[exp(-cs) F(s)] = u(t - c) f(t - c)
where F is the Laplace transform of f, and u(t) is the unit step function
To apply this property, we first identify c = 1 and F(s) = 1/(s² + 4), whose inverse transform is
L⁻¹[F(s)] = 1/2 L⁻¹[2/(s² + 2²)] = 1/2 sin(2t)
Then we find
L⁻¹[Y] = exp(2t - 2) u(t - 1) • 1/2 sin(2 (t - 1))
and so we end up with
y = 1/2 exp(2t - 2) u(t - 1) sin(2t - 2)
The function in the graph has the name of square function.
The domain of a function is all values of x the function can have. The domain of this function is all real numbers:
The range of a function is all values of y the function can have. The range of this function is all positive numbers, including zero:
In order to graph f(x) + 2, we just need to translate the graph 2 units up. To find the new points, we need to increase all y-coordinates by 2:
(-2, 6), (-1, 3), (0, 2), (1, 3), (2, 6)
Domain: {x ∈ ℝ}
Range: {y ∈ ℝ | y ≥ 2}
Then, in order to graph f(x) - 2, we just need to translate the graph 2 units down. To find the new points, we need to decrease all y-coordinates by 2:
(-2, 2), (-1, -1), (0, -2), (1, -1), (2, 2)
Domain: {x ∈ ℝ}
Range: {y ∈ ℝ | y ≥ -2}
