If f is defined by $f(x)= x^3 +2x^2 + x$ , how do you find the value of x when the average rate of change of f on the interval $x=-1$ to $x =2$ is equal to the instantaneous rate of change on the interval from $x=-1$ to $x=2$ ? The answer is x=0.786?

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If f is defined by $f(x)= x^3 +2x^2 + x$ , how do you find the value of x when the average rate of change of f on the interval $x=-1$ to $x =2$ is equal to the instantaneous rate of change on the interval from $x=-1$ to $x=2$ ? The answer is x=0.786?

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Answer 1 · 2015-04-06T13:20:11

Ok, this is a good one!
Consider that the instantaneous rate of change is the derivative, $f'(x)$ , while the average is the difference between two values (evaluated in $x=a and x=b$ ) of your function in an interval divided by the interva ( $b-a$ )l!!!
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If f is defined by $f(x)= x^3 +2x^2 + x$ , how do you find the value of x when the average rate of change of f on the interval $x=-1$ to $x =2$ is equal to the instantaneous rate of change on the interval from $x=-1$ to $x=2$ ? The answer is x=0.786?

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If f is defined by f(x)= x^3 +2x^2 + xf(x)= x^3 +2x^2 + x, how do you find the value of x when the average rate of change of f on the interval x=-1x=-1 to x =2x =2 is equal to the instantaneous rate of change on the interval from x=-1x=-1 to x=2x=2? The answer is x=0.786?

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If f is defined by $f(x)= x^3 +2x^2 + x$ , how do you find the value of x when the average rate of change of f on the interval $x=-1$ to $x =2$ is equal to the instantaneous rate of change on the interval from $x=-1$ to $x=2$ ? The answer is x=0.786?