How do you find the instantaneous rate of change for f(x)=x2−2x−1 for x=2?

Question

dabCrupedeedaejrg · Accepted Answer

We have to take the derivative of the function f(x), and then plug in x=2 into f'(x).

f (x) = \frac{x^{2} - 2}{x - 1}

So, we have to use the quotient rule, which states that,

\frac{d}{d x} (\frac{u}{v}) = \frac{u^{'} v - u v^{'}}{v^{2}}

Here,

u = x^{2} - 2, v = x - 1

∴ f^{'} (x) = \frac{2 x (x - 1) - 1 \cdot (x^{2} - 2)}{{(x - 1)}^{2}}

f^{'} (x) = \frac{2 x^{2} - 2 x - x^{2} + 2}{{(x - 1)}^{2}}

f^{'} (x) = \frac{x^{2} - 2 x + 2}{{(x - 1)}^{2}}

Plugging in x=2, we get

f^{'} (2) = \frac{2^{2} - 2 \cdot 2 + 2}{{(2 - 1)}^{2}}

f^{'} (2) = \frac{4 - 4 + 2}{1^{2}}

f^{'} (2) = \frac{2}{1}

f'(2)=2
So, the instantaneous rate of change will be 2.

YAMAGAMA699k · Accepted Answer

By finding the derivativ of f(x)f(x)=x2−2x−1f′(x)=(x2−2)⋅1−(x−1)⋅2xx−1Now we determinate the value of the divertive at the point x=2f&#039;(2)=2

How do you find the instantaneous rate of