<p>I really don't understand how to solve this problem</p>
<p>Suppose f(x) = {cos x , 0≤ x ≤ pi/3
{ax + b, pi/3 < x ≤ pi <<=piecewise function</p>
<p>a) Find a, and b so that f is differentiable at x = pi/3</p>
<p>how do you use the properties of continuity and differentiability to find the values of a and b????</p>
<p>first, cos(pi/3)=1/2, so the line y=ax+b must go through (pi/3, 1/2)</p>
<p>The derivative of cosx evaluated at pi/3 is -sin(pi/3)=-radical(3)/2</p>
<p>Thus the derivative of ax+b must be -radical(3)/2</p>
<p>The derivative of ax+b is simply a, so a=-radical(3)/2</p>
<p>Plug that in for a and the point (pi/3, 1/2) for x and y: </p>
<p>1/2=(-radical(3)/2)(pi/3)+b</p>
<p>I hope you can solve for be from there, because I don't want to.</p>
<p>In order for something to be differentiable, it must first be continuous, which is why I forced the line y=ax+b to pass through that specific point in the first step. Then, the derivative at a point must be the same when approaching from the right and from the left, which is why I set the derivative of ax+b (which is just a) to the derivative of cosx evaluated at pi/3.</p>
<p>I hope this helps, and that it is correct. Why are you doing calculus over the summer?</p>
<p>Sorry about the sloppy notation.</p>
<p>okay well if its differentiable, then its continuous. start by taking the derivs of both eqs. (-sinx and a). plug in th value of x (pi/3) and set the 2 derivs equal to each other. solve for a, getting a= -sq.root of 3 divided by 2. then go back and set the original eqs. equal to each other and fill in the x value and the new value for a. then just solve for b. you should get b=((sq.rt.3)pi+3)/6.</p>
<p>Okay, for piecewise functions to be differentiable, two things must be fulfilled.
1. They must be equal at the point where the first piece ends and the second piece begins, in this case, at x= pi/3
2. Their derivatives must be equal at the said point above.</p>
<p>So that means cos x = ax+b at x = pi/3, and also
-sin x = a when x = pi/3</p>
<p>I think you can do it from here, but I'll post the answer in case you have any more questions</p>
<p>-sin (pi/3) = a, so a = -sqrt(3)/2
cos (pi/3) = 0.5, so
0.5 = ax + b
we found a, so
b = 0.5 - ax
b = 0.5 - (-sqrt(3)/2)(pi/3)</p>
<p>wow thanks everyone...calc is a little hard for me to learn by myself.</p>
<p>"Why are you doing calculus over the summer?"</p>
<ul>
<li>i don't want to take two years of calc at school so i'm hoping to learn the AB curriculum and take BC next year</li>
</ul>
<p>just take calc at a community college...i did..</p>
<p>Even better would just be to take the AB class and study for the BC Test...
It's important to have a solid foundation in calculus if you're going to continue taking math. I would prefer to have an instructor teach me the basics. It shouldn't be too hard to learn the additional 3 chapters in BC calculus.</p>
<p>lildude, infinite series are difficult to master without some help. good luck.</p>