Use a linear approximation (or differentials) to estimate the given number
(2.001)^5
To estimate the value of (2.001)^5 using linear approximation, we can use differentials, specifically the concept of differentials in calculus. The differential of a function represents the change in the function value due to a small change in the input.
To start, let's define the function f(x) = x^5. Using differentials, we can write the change in f, Δf, as the derivative of f with respect to x, multiplied by the change in x, Δx:
Δf = f'(x) * Δx
Now, suppose we want to approximate the value of f(x) at a nearby point x + Δx, denoted as f(x + Δx). Using differentials, we can rewrite this as:
f(x + Δx) = f(x) + Δf
Substituting the definition of Δf from above:
f(x + Δx) = f(x) + f'(x) * Δx
In our case, f(x) = x^5 and x = 2.001. Let's choose a small Δx such as 0.001. Plugging in these values:
f(x + Δx) = f(2.001 + 0.001) = f(2.002) = (2.002)^5
and
f(x) = f(2.001) = (2.001)^5
To approximate the value of (2.001)^5, we can use the linear approximation formula:
f(x + Δx) ≈ f(x) + f'(x) * Δx
In our case, it becomes:
(2.002)^5 ≈ (2.001)^5 + f'(2.001) * 0.001
To evaluate f'(2.001), we can find the derivative of f(x) = x^5 with respect to x:
f'(x) = 5x^4
Evaluate f'(2.001):
f'(2.001) = 5(2.001)^4
Finally, substitute the values back into the linear approximation formula:
(2.002)^5 ≈ (2.001)^5 + 5(2.001)^4 * 0.001
Calculating this expression will give an estimate for (2.001)^5 using linear approximation.