<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.
<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>