use differentials to approximate the change in f(x) if x changes from 3 to 3.01 and f(x) = (3x^2-26)^10
derivative: 10(3x^2-26)^9
f(x+ delta x)= 1 + 10(3x^2-26)^9 * 0.01
= 1 + 10(0.01) <-- Substitute 3 for x
= 1 + 0.1
= 1.1
delta y= 1.1 - 1
delta y= 0.1
but 0.1 is incorrect. why so?
derivative is 10(3x^2-26)^9 * (6x)
The Chain Rule- of course! Thank you very much Steve. :-)
Apologies for the mistake in the calculation. Let's go through the steps again to find the correct approximation.
Given: f(x) = (3x^2 - 26)^10
1. Find the derivative of f(x):
f'(x) = 10(3x^2 - 26)^9
2. Use the differential approximation formula to approximate the change in f(x):
Δf(x) ≈ f'(x) * Δx
Here, Δx = 3.01 - 3 = 0.01
Δf(x) ≈ f'(x) * 0.01
Substitute x = 3 into f'(x):
f'(3) = 10(3(3)^2 - 26)^9
= 10(63 - 26)^9
= 10(37)^9
Δf(x) ≈ 10(37)^9 * 0.01
Please note that the value of Δf(x) will depend on the exact value of 37^9, which is a large number. It is recommended to use a calculator or a programming language to obtain an accurate approximation of the expression.
Let me know if I can assist you with anything else.
To find the approximate change in f(x) when x changes from 3 to 3.01 using differentials, we can use the formula:
Δf ≈ f'(x)Δx
First, let's find the derivative of f(x) = (3x^2-26)^10 with respect to x.
The derivative is given by:
f'(x) = 10(3x^2-26)^9 * d/dx(3x^2-26)
Now, let's find the value of f'(x) when x = 3:
f'(3) = 10(3(3)^2-26)^9
= 10(27-26)^9
= 10(1)^9
= 10
Next, we substitute into the formula Δf ≈ f'(x)Δx:
Δf ≈ f'(3) * Δx
Given that Δx = 3.01 - 3 = 0.01, we have:
Δf ≈ 10 * 0.01
= 0.1
So, the approximate change in f(x) when x changes from 3 to 3.01 is indeed 0.1, not 0.01. The value you calculated, 0.01, is the value of Δx, which represents the change in x. The actual change in f(x) is given by Δf.