Solution

(a) The first form of the difference quotient usually stumps students. You have to plug that whole quantity (x plus delta x) into the function f(x), in place of the x. In other words, wherever an x appears in f(x), replace it with (x plus delta x). Then, you have to subtract from that f(x) itself. Don't forget to distribute that negative through all of f(x)

Be careful with your notation. Now, factor out the delta x in the numerator and cancel it with the matching term in the denominator.

Now, replace delta x with 0 in order to evaluate the limit, and plug in 2 to the derivative to answer the question posed by the problem..

The second form of the difference quotient is a little easier to handle, but the factoring it involves is a little trickier. Start by plugging into the formula end evaluating f(2) as the formula requires.

In order to be able to complete this problem (i.e. to plug in 2 for x and not have a heart attack because of 0 in the denominator), we need the numerator to also have a factor of (x - 2) so that the denominator cancels out. I like synthetic division to factor out ugly polynomials when you know a root, so I have elected to use it.

Now, we can write the ugly cubic polynomial in factored form and cancel out the troubling denominator and we get the correct answer. Notice that we don't have to go back and plug in 2 like in the previous difference quotient, but we also do not get the generic derivative like we did in the first either.

(b) To approximate the derivative geometrically, you must first know what the derivative is--the slope of the tangent line to the function at the given point. The best way to approximate a tangent line is to use a secant line with two x values which are relatively close together. The line which passes through the function's values at x = 1.95 and x = 2 has a slope nearly equal to that of the tangent line to f through x = 2. An x value of 1.96 or even 1.99999 would be even closer, but let's not go crazy...we already know what the real answer is for crying out sakes. Through a quick calculation we get . Use these values and the definition of slope from Algebra I (change in y divided by change in x) to get the secant slope.

The answer of 19.6025 is close to 20, but not all that close. To get the exact answer of 20, we had to use a value of x closer to 2 than 1.95. Even 1.99 will not be close enough to be exact. The whole design of the difference quotient is to reach that value of x so close to the target value that it basically is the target value, in this case 2. When that happens, the answer is exact. There is no single real number close enough to 2 that gives you that exact value, which is why the method described in (b) can be an approximation at best.

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