Best solutions to - "What is the instantaneous rate of change at..."

Jerimiah Boone

Answered question

2022-03-17

What is the instantaneous rate of change at x=2 of the function given by

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

Jarrett Chan

Beginner2022-03-18Added 6 answers

Let

g (x) = x^{2} - 1

and h(x)=x-1
so

f (x) = \frac{g (x)}{h (x)}

By the quotient rule for derivatives

f^{'} (x) = \frac{g^{'} (x) \cdot h (x) - g (x) \cdot h^{'} (x)}{{(h (x))}^{2}}

g'(x)=2x
h'(x)=1
So,

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

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

The instantaneous rate of change at x=2 is therefore

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

= \frac{2}{1} = 2

ngiqulelexvg

Beginner2022-03-19Added 4 answers

What is the instantaneous rate of change at x=2 of the function given by

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

?
Using the definition:

lim_{x \to 2} \frac{f (x) - f (2)}{x - 2}

= lim_{x \to 2} \frac{\frac{x^{2} - 2}{x - 1} - \frac{2^{2} - 2}{2 - 1}}{x - 2}

= lim_{x \to 2} \frac{\frac{x^{2} - 2}{x - 1} - 2}{x - 2}

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

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

= lim_{x \to 2} \frac{x^{2} - 2 - 2 x + 2}{(x - 2) (x - 1)}

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

= lim_{x \to 2} \frac{x (x - 2)}{(x - 2) (x - 1)}

= lim_{x \to 2} \frac{x}{x - 1}

= \frac{2}{2 - 1} = 2

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