Differentiability. Example Problem

Differentiability. Example Problem Differentiability is a measure of how smooth a function is. A function is differentiable at apoint if its derivative exists at that point and is non-zero. The graph of a differentiablefunction is infinitely thin in the sense that it can be covered by an arbitrarily smallrectangle. A simple way to think about differentiability is as follows: If you draw two tangent lines to a curve at two different points, and then draw all thepossible curves between these two tangent lines, then only those curves that pass throughcommon points on both of your tangent lines will intersect with both of them. If thisintersection happens at any point other than a maxima or minima, then that curve is said tobe differentiable at that point (and hence everywhere else). Given the graph of f of x and its derivative f ′(x), find values of a and b such that f(x) isdifferentiable. This can be very similar to the continuity we did earlier, where you need to use thedefinition of differentiability to determine if a function is differentiable or find those values. In order to show differentiability, you must show two things. First, you have to show that f iscontinuous at x=1. The second thing you have to show is that f'(x) exists at x=1, which wewill do below. Hopefully, you have plenty of room on your paper. To demonstrate continuity, recall the definition of continuity: The limit as x approaches 1from the left of f(x) equals the limit as x approaches 1 from the right of f(x). If this is multiple choice, that's a different story, but if it's for your response, you shouldshow how you derived the function value at one. If limit x approaches 1 from the left, then x is less than 1. The value of the limit is equal toax squared minus 4b plus 2. Evaluate: a minus 4b plus 2. When x approaches 1 to the right, or increases in value, the value of bx – 2a approaches b– 2a. Evaluated when x is 1, or equals 1, you will have b minus 2a. And formally, this expression equals one divided by f of 1. We will assume this to be true ifwe can show that f of 1 is equal to the sum of f and 1/f. From here, all you have to do is manipulate the equation. You can say that 3a = 2a + 2,move the 4b over to both sides of the equation, and then subtract 5b from both sides. It’sall algebra. Since we require two equations to solve for two unknowns, where can we find the secondequation? The answer is f prime. To show that f prime exists, one can use the limit theorem to show that as x approaches 1from the left and right sides of a derivative, the functions become equal. I will not use the alternate definition of derivative here because once I know the power rule,I do not need the definition of derivative to show a derivative. I can take the derivative of

this expression and be just fine. So let us consider x approaches 1 from the left; thuslimiting the left hand side to zero. And so the derivative of this would be the left-handderivative. The derivative of ax squared is 2ax. The derivative of 4b is 0 because b is a constant.There are two constants in the equation, so the derivative of each constant is 0. Thatmeans that 2ax multiplied by 2a equals 0. The limit of bx as x approaches 1 from the right is the derivative of bx, which is b minus thederivative of 2a squared x. When x is equal to 1, this equals ax to the negative one-half. Evaluate it when x equals 1 and you get b minus a. Therefore, the components of f prime must be equal to each other. Therefore, 3a is equalto b. Now I have two linear equations and two variables, so I can solve this problem by usingthe substitution method. Let's say a = 3. Then 3a + 2 = 5(3)a. Now we're looking for the value of a that makes thisequation true. Subtract 3 from both sides, so that would be 12; now divide both sides by12, so we get a=1/6. If we know a, we can find b. Looking over here three times and that's 1/2 is the value for b.So now we know the values of a and b that make f of x continuous and differentiable.