Use a linear approximation to estimate the number (27.07)^(2/3)

User Icon for Reiny Reiny answered
7 years ago

using:

f(x)≈f(xo)+f′(xo)(x−xo) <-- should be in your text

27.07 = 27 + .07

let f(x) = x^(2/3)
f'(x) = (2/3)x^(-1/3)

let xo = 27, x = 27.07
f(27.07) ≈ f(27) + f'(27)(27.07 - 27)
= 27^(2/3) + (2/3)(27)^(-1/3) (.07)
= 9 + (2/3)((1/3)(.07)
= 9 + .01555..
= 9.01555..

by calculator, 27.07^(2/3) = 9.0155488..
not bad !!

User Icon for Clown Bot Clown Bot answered
1 year ago

Well, let's give it a shot! We can use a linear approximation to estimate the value of (27.07)^(2/3).

To do that, we'll start by finding the derivative of the function f(x) = x^(2/3). The derivative of this function is 2/3 * x^(-1/3).

Now, let's choose a number close to 27.07 (let's say x = 27), and find both f(27) and f'(27).

f(27) = 27^(2/3) ≈ 15.196

f'(27) = (2/3) * (27^(-1/3)) ≈ 0.222

Using these values, we can form a linear approximation.

The linear approximation formula is:

L(x) = f(a) + f'(a) * (x - a)

Using the chosen values:
L(27.07) ≈ f(27) + f'(27) * (27.07 - 27)

L(27.07) ≈ 15.196 + 0.222 * 0.07

L(27.07) ≈ 15.196 + 0.01554

L(27.07) ≈ 15.21154

Therefore, a linear approximation estimates that (27.07)^(2/3) is approximately 15.21154.

But remember, this is just an approximate value! It's always a good idea to double-check.

User Icon for Step-by-Step Bot Step-by-Step Bot answered
1 year ago

To estimate the value of (27.07)^(2/3) using linear approximation, we'll start by finding the equation of the tangent line to the curve at a nearby point.

Let's start by finding a nearby point that we can use as a starting point for our linear approximation. One possibility is to use 27, which is the closest perfect cube to 27.07. The cube root of 27 is 3, so we can use this point (27, 3) as our starting point.

Next, we need to find the slope of the tangent line at this point. We can do this by finding the derivative of the function f(x) = x^(2/3) and evaluating it at x = 27.

The derivative of f(x) = x^(2/3) can be found using the power rule. We have:

f'(x) = (2/3)x^(-1/3)

Evaluating this at x = 27:

f'(27) = (2/3)(27)^(-1/3) = (2/3)(1/3) = 2/9

Now we have the slope of the tangent line at the point (27, 3), which is 2/9.

Finally, we can use the point-slope form of a linear equation to write the equation of the tangent line:

y - 3 = (2/9)(x - 27)

Now, we can use this tangent line equation to estimate the value of (27.07)^(2/3).

Plug in x = 27.07 into the equation and solve for y:

y - 3 = (2/9)(27.07 - 27)
y - 3 = (2/9)(0.07)
y - 3 = 0.0156

Adding 3 to both sides, we get:

y = 3.0156

Therefore, using linear approximation, we can estimate that (27.07)^(2/3) is approximately 3.0156.

User Icon for Explain Bot Explain Bot answered
1 year ago

To use linear approximation to estimate the value of (27.07)^(2/3), we'll start by finding a point close to 27.07 where we know the exact value of the function. Let's use 27, which is a perfect cube, so we know that (27)^(2/3) = 9.

Next, we'll find the derivative of the function f(x) = x^(2/3) with respect to x. The derivative of f(x) is f'(x) = (2/3)x^(-1/3).

Now, we can use linear approximation to estimate the value of (27.07)^(2/3) near x = 27. We'll use the formula:

f(x) ≈ f(a) + f'(a)(x - a)

where f(a) is the known value of the function at x = a, f'(a) is the derivative of the function at x = a, and (x - a) is the difference between the desired x-value and a.

Using a = 27, f(a) = 9, and f'(a) = (2/3)(27)^(-1/3), we have:

(27.07)^(2/3) ≈ 9 + [(2/3)(27)^(-1/3)](27.07 - 27)

Calculating this, we get:

(27.07)^(2/3) ≈ 9 + (2/3)(27)^(-1/3)(0.07)

Simplifying further, we get:

(27.07)^(2/3) ≈ 9 + (2/3)(1/3)(0.07)

(27.07)^(2/3) ≈ 9 + (2/9)(0.07)

Calculating this, we find:

(27.07)^(2/3) ≈ 9 + 0.015555...

Therefore, the linear approximation of (27.07)^(2/3) is approximately 9.015555...