Calc bc plz help me

<p>I get inflection points but i dont get how to do this problem. Determine a so that the function f(x)=x^2+(a/x) has an inflection point at x=1</p>

<p>Well, assuming that the variable a is constant, how would you find the inflection point(s) for the above function?</p>

<p>Once you’ve found that solution for x in terms of a, then you need to figure out what a would have to be so that x = 1 would be an inflection point…</p>

<p>What TheMathProf said. Solve it as if you were solving any ordinary inflection points problem. Once you get your second derivative, set it equal to zero, and plug in x, you’ll realize everything lines up perfectly. :)</p>

<p>If dy/dx = sqrt x, then the average rate of change of y with respect to x on the closed interval (0,4) is ?</p>

<p>the options given are 1/16, 1, 4/3, sqrt2, 2</p>

<p>How do I solve this?</p>

<p>I do not seem to get any of the above options as my answer.</p>

<p>Help please.</p>

<p>average rate of change = [f(b) - f(a)]/(b - a)
F(x)=(2/3)X^(3/2) + C
F(0)=C
then
F(x)=(2/3)X^(3/2)+F(0)
Then<br>
F(0)= F(0)
F(4)= (16/3)+F(0) </p>

<p>Plug it into the equation [f(b) - f(a)]/(b - a)
[(16/3)+F(0) - F(0)] / 4-0 </p>

<p>= 4/3</p>