Compute approximate value of a function change
$y=x^3 - x^2$ when $x=2$ and $\Delta x=0.01$. Any ideas how to solve it?
My idea
Let $Δx = dx$ and $Δy= dy$, then $f(x+Δx) = f(x) + f′(x)\timesΔx, \quad Δx = 2-0.01=1.99$, then $f(0,01) + f ′ (0,01) \times 1,99 = \ldots$
Am I right?
You get lost when you get to $\Delta x = 2-0.01=1.99$.
You are given that $\Delta x = 0.01$. It doesn't suddenly change to $1.99$.
The key realization is that $$f'(x) =\frac{dy}{dx} = \lim_{\Delta x \to 0}\frac{\Delta y}{\Delta x}$$ so for small values of $\Delta x, f'(x) \approx \frac{\Delta y}{\Delta x}$, and therefore $$f(x + \Delta x) - f(x) = \Delta y \approx f'(x)\Delta x\\f(x + \Delta x) \approx f(x) + f'(x)\Delta x$$
Now you are given that $x = 2, \Delta x = 0.01$ and $y = f(x) = x^3 - x^2$. So: $$f(2) = 8 - 4 = 4\\f'(x) = 3x^2 - 2x\\f'(2) = 12 - 4 = 8$$
Plugging that in: $$f(2 + 0.01) \approx 4 + 8 \times 0.01\\f(2.01) \approx 4.08$$
The actual value is $f(2.01) = 4.080501$, so the approximation is pretty good.