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