Find the linear approximation of the function f(x)=^3sqrt(1+3x) at a=0 and use it to approximate ^3sqrt(1.03)Is this an overestimate or underestimate of the actual value? Explain your answer in terms of the tangent line and the curve.
if ^3sqrt means cube root, then
f(x) = (1+3x)^(1/3)
df/dx = (1+3x)^(-2/3)
so, df = (1+3x)^(-2/3) dx
At x=0, df = dx
So, since dx=1.03, f=f(0)+1.03=1.03
at x=0, the line y=x+1 is tangent to the curve
I think you meant to approximate by using x=1. In that case,
f(1) = ∛4 = 1.587
df = 1/∛16 dx
since dx=.03, df=.03/∛16 = .015/∛2 = .0119
f(1.03) = 1.587+0.0119 = 1.599
at x=1, the tangent line is
y-1.587 = .3969(x-1) or
y = .3969x + 1.19
Visit http://rechneronline.de/function-graphs
and enter the three functions as
(1+3x)^(1/3)
x+1
.3969x + 1.19
then click "Draw" and you will see how the lines approximate the curve.
at x=0, since f(0) = 1, the
approximation at x=1.03 would be
f(1.03) = 1+1.03 = 2.03
In f(1.03) = 1.587+0.0119 = 1.599, did you apply the linear approximation equation L(x)=f(a)+f'(a)(x-a)? Because I'm still a little confused on how to apply that equation to problems like these.
That's exactly what I did.
f(1) = 1.587
f'(1) = .3968
(1.03-1) = .03
To find the linear approximation of a function at a specific point, we can use the formula:
L(x) = f(a) + f'(a)(x - a),
where f(a) is the value of the function at a, f'(a) is the derivative of the function evaluated at a, and (x - a) is the difference between the given point x and the point a.
Given the function f(x) = ∛(1 + 3x), we want to find the linear approximation at a = 0.
1. Calculate f(0):
f(0) = ∛(1 + 3(0))
= ∛1
= 1.
2. Calculate f'(x):
f'(x) = d/dx ∛(1 + 3x)
= 1/3 * (1 + 3x)^(-2/3).
3. Evaluate f'(0):
f'(0) = 1/3 * (1 + 3(0))^(-2/3)
= 1/3 * 1^(-2/3)
= 1/3.
Now we have the required values to find the linear approximation L(x) at x = 0:
L(x) = f(0) + f'(0)(x - 0)
= 1 + (1/3)(x)
= 1/3x + 1.
To approximate ∛(1.03) using the linear approximation, substitute x = 0.03 into L(x):
L(0.03) = 1/3(0.03) + 1
= 0.01 + 1
= 1.01.
To determine whether this is an overestimate or underestimate, we need to compare it with the actual value of ∛(1.03).
The tangent line approximation using L(x) represents a linear approximation of the original curve at the given point a. In this case, a = 0. The tangent line intersects the curve at a specific point, and the linear approximation value will be either above (overestimate) or below (underestimate) the actual curve value.
Since L(0.03) = 1.01 and ∛(1.03) ≈ 1.0098, the linear approximation is an overestimate because L(0.03) is slightly larger than the actual value of ∛(1.03).