<p>That's the problem. Gotta verify that the function satisfies the mean value theorem on the interval. then find numbers that satisfy the MVT.</p>
<p>The derivative of the function is (sqr root x) -1 / 2x sqr root x . And i've gotten that that equals -1/6. From here the algebra is killing me! Am I doing something wrong? Please help!!!</p>
<p>f'(x) = [ f(b) - f(a) ] / (b - a)</p>
<p>(-1/2)x^(-3/2) = [ 1/2 - 1 ] / 3</p>
<p>-1 / (2x^3/2) = -1/6</p>
<p>2x^3/2 = 6</p>
<p>x^3/2 - 3 = 0</p>
<p>(x^3/4 + sqrt3)(x^3/4 - sqrt3) = 0</p>
<p>negative can't exist so only value is:</p>
<p>x^3/4 = sqrt 3</p>
<p>to make it easier square values
X^3/2 = 3</p>
<p>both to the 2/3 power</p>
<p>x = cube root of 9</p>
<p>approx. x = 2.1</p>
<p>Hence a c value, ~2.1, lies w/in the interval [1,4].</p>
