<p>I was reviewing some problems and I kept getting stuck on this problem. Can anyone help me out here.</p>
<p>A function f is defined by f(x)=x^2-3*sqrt(x) for the interval [1,4]. During this interval, at what value is the instantaneous rate of change of f equal to the average rate of change over the interval [1,4]</p>
<p>Thanks!</p>
<p>bump anyone know the answer…</p>
<p>plzzzz help</p>
<p>Its when x is about 2.477</p>
<p>The instantaneous rate of change is the definition of the derivative so in this case f '(x). Avergae rate of change is [f(b)-f(a)]/ (b-a).</p>
<p>Set them both equal to each other:</p>
<p>2x - 3/(2 * x^1/2) = [f(4) - f(1)] / 3 —> 2x - 3? 2sqrt(x) = 4—> x = ~2.477</p>
<p>The value of x is 2.477, the value of the instantaneous rate of change is 4; not sure which one you want :)</p>