How do you find the instantaneous rate of change of f(x)=x2−2x+4 at x=-1?

Question

CyncgotoCancey1k6 · Accepted Answer

When you calculate a function&#039;s derivative, you obtain an other function representing the variations of the first function&#039;s curve&#039;s slope.A curve&#039;s slope is the instantaneous variation rate of the curve&#039;s function at a given point.Therefore, if you are looking for the instantaneous variation rate of a function at a given point, you should calculate this function&#039;s derivative at said point.In your case:f(x)=x2−2x+4→ variation rate at x=-1?Calculating the derivative:f′(x)=d(x2)dx−d(2x)dx+d4dx=2x−(−2x2)+0=2x+2x2Now, you just need to replace x in f&#039;(x) with its given value, x=-1f′(−1)=2(−1)+2(−1)2=−2+2=0The derivative is null, therefore the instantaneous change rate is null and the function doesn&#039;t increase or decrease at this specific point.

pobijedi6wro · Accepted Answer

The instantaneous rate of change of the function f(x) at the point

x = x_{0}

is the derivative of the function f(x) evaluated at the point

x = x_{0}

.
This means that we need to find the derivative of

x^{2} + 4 - \frac{2}{x}

and evaluate it at x=-1.
So, find the derivative of the function:

\frac{d}{d x} (x^{2} + 4 x - \frac{2}{x}) = 2 x + \frac{2}{x^{2}}

Finally, evaluate the derivative at x=-1.

(\frac{d}{d x} (x^{2} + 4 - \frac{2}{x})) {|_{{(x = - 1)}} = (2 x + \frac{2}{x^{2}}) |}_{(x = - 1)} = 0

Therefore, the instantaneous rate of change of

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

at x=-1 is 0.

How do you find the instantaneous rate of

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