Question #8a726

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Question #8a726

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Answer 1 · 2018-01-08T10:32:22

See below.

Explanation:

To find the instantaneous rate of change, we need to find the derivative of $1/x$ . Putting in values of $x$ will give us the gradient of the tangent at that point. This is different from the average rate of change, which is the gradient of the secant line that joins $2$ points.

So the instantaneous rate of change is given by $f'(a)$ , and the average rate of change on an interval $[a,b]$ is given by

$((f(b))-(f(a)))/(b-a)$

$1/x=x^-1$

Using the power rule:

$dy/dx(x^n)=nx^(n-1)$

$dy/dx(x^-1)=-x^(-2)=-1/x^2$

$:.$

$x_0=2$

$f'(x_0)=-1/(2)^2=color(blue)(-1/4)$

$x_1=3$

$f'(x_1)=-1/(3)^2=color(blue)(-1/9)$

Instantaneous Rate of Change:

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Average Rate of Change:

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Question #8a726

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