Use the Linear Approximation to estimate ∆f=f(3.5)−f(3) for f(x)=2/(1+x^(2))?
Δf ≈
Estimate the actual change.
(Use decimal notation. Give your answer to five decimal places.)
Δf=
Compute the error and the percentage error in the Linear Approximation.
(Use decimal notation. Give your answer to five decimal places.)
Error =
linear approximation means get the slope.
df/dx=-2(2x)/(1+x^2)^2
at at x=3.25, df/dt= -4*3.25/(1+3.25^2)^2=-.0972
df=-.0972*x but x= 3.5-3=.5
df=-.0488
actual
f(3)=2/(10)=1/5
f(3.5)=2/(1+3.5^2)=.150
f(3.5-f(3)=-.050
error (.0488-.050)/.050 about 2 percent
f(3.5)-f(3)=-.350
it says the answer is incorrect
To estimate ∆f=f(3.5)−f(3) for f(x)=2/(1+x^(2)), we can use the linear approximation.
First, let's find the derivative of f(x) with respect to x:
f'(x) = d/dx (2/(1+x^(2)))
Using the quotient rule, we have:
f'(x) = (-2x) / (1+x^(2))^(2)
Next, let's find the equation of the tangent line at x=3:
Using the point-slope form of a line, y - y1 = m(x - x1), where m is the slope and (x1, y1) is a point on the line, we have:
y - f(3) = f'(3)(x - 3)
Substituting the values, we get:
y - f(3) = (-2(3)) / (1+(3)^(2))^(2) * (x - 3)
Simplifying, we have:
y - f(3) = -6 / (1+9)^(2) * (x - 3)
y - f(3) = -6 / 10 * (x - 3)
y - f(3) = -0.6 * (x - 3)
y = -0.6x + 1.8 + f(3)
Now, let's find f(3.5) using the linear approximation:
f(3.5) ≈ -0.6(3.5) + 1.8 + f(3)
f(3.5) ≈ -2.1 + 1.8 + f(3)
f(3.5) ≈ -0.3 + f(3)
Now, let's compute f(3) to find the actual change:
f(3) = 2/(1+(3)^(2))
f(3) = 2/(1+9)
f(3) = 2/10
f(3) = 0.2
Therefore, Δf ≈ f(3.5) - f(3) ≈ (-0.3 + f(3)) - (0.2)
Δf ≈ -0.3 + f(3) - 0.2
Δf ≈ -0.3 + 0.2
Δf ≈ -0.1
The estimate for the actual change is Δf = -0.1
Next, let's compute the error in the linear approximation.
The error is given by the difference between the actual change and the linear approximation:
Error = Δf - (f(3.5) - f(3))
Substituting the values, we have:
Error = -0.1 - (-0.3 + 0.2)
Error = -0.1 + 0.3 - 0.2
Error = 0
The error in the linear approximation is 0.
Finally, let's compute the percentage error:
Percentage Error = (Error / Δf) * 100
Substituting the values, we have:
Percentage Error = (0 / -0.1) * 100
Percentage Error = 0
The percentage error in the linear approximation is 0%.
To use the Linear Approximation to estimate Δf = f(3.5) - f(3) for f(x) = 2/(1+x^2), we first need to find the linear approximation of the function.
The linear approximation of f(x) at a given point a is given by the equation:
L(x) = f(a) + f'(a)(x - a)
where f'(a) is the derivative of f(x) evaluated at a.
First, let's find the derivative of f(x):
f(x) = 2/(1+x^2)
f'(x) = -4x/(1+x^2)^2
Now, let's evaluate f(3) and f'(3):
f(3) = 2/(1+3^2) = 2/10 = 0.2
f'(3) = -4*3/(1+3^2)^2 = -12/100 = -0.12
Now we can use the linear approximation formula to estimate Δf:
L(x) = f(a) + f'(a)(x - a)
L(3.5) = f(3) + f'(3)(3.5 - 3)
L(3.5) = 0.2 + (-0.12)(3.5 - 3)
L(3.5) = 0.2 - 0.12(0.5)
L(3.5) = 0.2 - 0.06
L(3.5) = 0.14
So our estimate for Δf is approximately 0.14.
To estimate the actual change, we subtract f(3) from f(3.5):
Δf = f(3.5) - f(3)
Δf = 0.14 - 0.2
Δf = -0.06
The actual change in f is approximately -0.06.
To compute the error in the linear approximation, we subtract the linear approximation from the actual change:
Error = Δf - L(3.5)
Error = -0.06 - 0.14
Error = -0.2
To compute the percentage error, we divide the error by the actual change and multiply by 100:
Percentage Error = (Error / Δf) * 100
Percentage Error = (-0.2 / -0.06) * 100
Percentage Error = 333.33333
So the error in the linear approximation is approximately -0.2 and the percentage error is approximately 333.33333%.