Help on calculus question!

Test Preparation
1

<p>This is from the 2003 free responses #6c:</p>

<p>Suppose the function g is defined by:</p>

<p>[k*squareroot(x+1)] for x: [0,3]
and [mx+2] for x: (3, 5]</p>

<p>where k and m are constants. If g is differentiable at x = 3, what are the values of k and m?</p>

<p>Help please!</p>

2

<p>well, for it to be differentiable at x=3, the derivative of both functions must be equal at that point...</p>

3

<p>Wow (duh!) why didn't I notice that?! Thanks!</p>

4

<p><a href="http://apcentral.collegeboard.com/repository/ap03_sg_calculus_ab_26472.pdf%5B/url%5D"&gt;http://apcentral.collegeboard.com/repository/ap03_sg_calculus_ab_26472.pdf&lt;/a&gt;&lt;/p>

5

<p>The limits also have to be equal. Just solving for the derivatives won't give you enough information to find the values of k and m.</p>

<p>The limit as x -> 3 from the left of the first function is 2k, and the limit as x -> 3 from the right of the second function is 2k = 3m + 2.</p>

<p>When you solve for the derivatives, you get (1/4)k = m. Now you can use substitution to solve for one of the constants, and then use that to solve for the other constant.</p>

6

<p>g(3) and g'(3) must be the same for both parts of the piecewise function, although g(3) and g'(3) are not equal.</p>

<p>g(x):
k((x+1)^(1/2)) = mx+2
k(4^1/2) = 3m+2
2k = 3m +2 </p>

<p>g'(x):
(k/2)((x+1)^(-1/2)) = m
(k/2)(4^-1/2) = m
(k/2)(1/2) = m
k = 4m</p>

<p>Thus, you have
2k = 3m + 2
k= 4m</p>

<p>2(4m) = 3m + 2
5m = 2
m=2/5, k=8/5</p>