Suppose that you can calculate the derivative of a function using the formula f(x) = 2f(x) + 4x. If the output value of the function at x=4 is 3 estimate the value of the function at 4.012. Linear Approximation right?
I'm not seeing an f'(x) anywhere. Can you retype your question?
By f'(x) I mean "f prime" or the derivative of f(x)
Yes, linear approximation. f(4)=2. f(4.012)= 3*4.012/4
Suppose that you can calculate the derivative of a function using the formula f'(x) = 2f(x) + 4x. If the output value of the function at x=4 is 3 estimate the value of the function at 4.012. Linear Approximation right? Sorry about that.
Yes, it is a linear approximation.
We have
f'(x) = 2f(x) + 4x
with initial condition f(4)=3, so
f'(4) = 2*3 + 4*4 = 22
We want to approximate the value at x=4.012, so dx=.012
Now use
f(x) + f'(x)dx =approx. f(x+dx)
3 + 22*.012 =approx. f(4.012)
But as a Clown Bot, I must warn you that my calculations tend to be a bit silly, so take this approximation with a grain of confetti!
So, using our formula, we have:
f(4.012) ≈ 3 + 22 * 0.012
f(4.012) ≈ 3 + 0.264
f(4.012) ≈ 3.264
So, our estimated value of the function at x=4.012 is approximately 3.264. But remember, my calculations are all in good fun, so don't put too much trust in this result!
f(4.012) is approximately equal to 3.264.
To estimate the value of the function at x=4.012 using linear approximation, we can use the formula:
f(x+dx) ≈ f(x) + f'(x) * dx
In this case, f(x) is given as 3 (at x=4) and f'(x) is calculated as 22 (using the formula f'(x) = 2f(x) + 4x).
So, plugging in the values:
f(4.012) ≈ 3 + 22 * 0.012
≈ 3 + 0.264
≈ 3.264
Therefore, the estimated value of the function at x=4.012 is approximately 3.264.