Use differentials to approximate the value of the expression.
cube root of 25
I set x=27 and dx=-2 and got ~3.0741, what did I do wrong?
our f(x) = x^(1/3) , and
f ' (x) = (1/3)x^(-2/3) = 1/(3(x)^(2/3) )
I will go with your x = 27 , dx = -2
f(x + dx) is appr = f(x) + f ' (x) (dx)
f(25) = f(27) + (1/3)(27)^(-2/3)
= 3 + (1/27)(-2)
= 3 - .0741
= 2.926
you simply added the part at the end instead of subtracting it. Looks like you did the rest ok
(I would give you most of the marks)
notice calculator answer is 2.924, so our approximation is pretty good.
Oh! Minor mistake, thanks for the help!
To approximate the value of the cube root of 25 using differentials, you need to follow the correct steps:
1. Start with the function that represents the cube root: f(x) = ∛x.
2. Find the derivative of f(x) with respect to x: f'(x) = 1/(3√(x^2)).
3. Choose a value near 25 to use as your approximation point. In this case, you can use x = 27.
4. Calculate the differential of f(x) as follows:
- df = f'(27)dx
- df = (1/(3√(27^2)))dx
- df = (1/9)dx
Note: dx represents the change in x, which is the difference between the value you chose (27) and the value you are approximating (25), so in this case, dx = 25 - 27 = -2.
5. Plug in the value of dx:
- df = (1/9)(-2)
- df = -2/9.
Now, to approximate the value of the cube root of 25, add the differential to the value you chose as your approximation point:
- Approximation = x + df
- Approximation = 27 + (-2/9)
- Approximation = 27 - 2/9
- Approximation ≈ 26.7778.
Therefore, the correct approximation of the cube root of 25 using differentials is approximately 26.7778. It seems you made a minor mistake when calculating the value of dx.
To approximate the value of the expression, the method you used is called differentials. However, there is a mistake in your calculations. Let's go through the process correctly.
First, we express the cube root of 25 as a function:
f(x) = (∛x)
Next, we can re-write this function as:
f(x) = x^(1/3)
Now, we can find the differential of the function using calculus. The differential of a function f(x) is denoted as df and can be approximated using the formula:
df ≈ f'(x) * dx
where f'(x) represents the derivative of f(x) with respect to x, and dx represents the change in x.
To find the derivative f'(x), we differentiate the function f(x) = x^(1/3) with respect to x:
f'(x) = (1/3) * x^(-2/3)
Now, substitute the given values x = 25 and dx = -2 into the differential formula:
df ≈ f'(x) * dx
≈ (1/3) * (25)^(-2/3) * (-2)
Calculating this expression allows us to approximate the value:
df ≈ (1/3) * (1/25^(2/3)) * (-2)
≈ -2/75
Finally, we can approximate the cube root of 25 as:
∛25 ≈ 25^(1/3) + df
≈ 25^(1/3) + (-2/75)
Now, we can calculate the approximation correctly:
∛25 ≈ 2.924087
Therefore, the correct approximation for the cube root of 25 using differentials is approximately 2.924087.
It seems that you made an error in calculating the derivative f'(x) and didn't substitute the correct values for x and dx into the differential formula.