Differentiability. Example Problem
Differentiability is a measure of how smooth a function is. A function is differentiable at a point if its derivative exists at that point and is non-zero. The graph of a differentiable function is infinitely thin in the sense that it can be covered by an arbitrarily small rectangle. 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 the possible curves between these two tangent lines, then only those curves that pass through common points on both of your tangent lines will intersect with both of them. If this intersection happens at any point other than a maxima or minima, then that curve is said to be 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) is differentiable. This can be very similar to the continuity we did earlier, where you need to use the definition 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 fis continuous at ×=1. The second thing you have to show is that f(x) exists at ×=1, which we will do below. Hopefully, you have plenty of room on your paper. To demonstrate continuity, recall the definition of continuity: The limit as x approaches 1 from 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 should show 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 to ax squared minus 46 plus 2. Evaluate: a minus 46 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 fof 1. We will assume this to be true if we can show that f of 1 is equal to the sum off 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’s all algebra. Since we require two equations to solve for two unknowns, where can we find the second equation? The answer is f prime To show that f prime exists, one can use the limit theorem to show that as x approaches 1 from 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; thus limiting the left hand side to zero. And so the derivative of this would be the left-hand derivative. The derivative of ax squared is 2ax. The derivative of 46 is O because b is a constant. There are two constants in the equation, so the derivative of each constant is 0. That means 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 the derivative 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 off prime must be equal to each other. Therefore, 3a is equal to b. Now I have two linear equations and two variables, so I can solve this problem by using the substitution method. Let’s say a = 3. Then 3a + 2 = 5(3)a. Now we’re looking for the value of a that makes this equation true. Subtract 3 from both sides, so that would be 12; now divide both sides by 12, 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 × continuous and differentiable. f(x)={ax^2 -4b+2, x<1; bx-2asqrtx, x=>1} [Table]
