170- Midterm 1

The exam is here.

Solutions follow.

Problem 1 asks to find \displaystyle \lim_{x\to 2}\frac{x^3-8}{x-2}.

There are several ways to proceed. The most direct consists in factoring the numerator, recalling the formula

a^3-b^3=(a-b)(a^2+ab+b^2).

In this case, we have x^3-8=x^3-2^3=(x-2)(x^2+2x+4), so

\displaystyle \lim_{x\to 2}\frac{x^3-8}{x-2}=\lim_{x\to2}\frac{(x-2)(x^2+2x+4)}{x-2} \displaystyle =\lim_{x\to2}x^2+2x+4=4+4+4=12.

Another way, which is more useful sometimes, is to recognize that the fraction is a derivative in disguise: Let’s write x=2+\Delta x, so that saying that x\to2 is the same as saying that \Delta x\to 0. Then x^3-8=(2+\Delta x)^3-2^3 and x-2=\Delta x. Then, letting f(x)=x^3, we have

\displaystyle \lim_{x\to2}\frac{x^3-8}{x-2}=\lim_{\Delta x\to0}\frac{f(2+\Delta x)-f(2)}{\Delta x}=f'(2).

Since f'(x)=3x^2, then f'(2)=3\times 4=12, just as before.

Problem 2 asks us to graph the derivative of the function given by the following graph (click to enlarge):

The graph of the derivative is shown below in blue (click to enlarge), together with the original curve:

To see that something like this shape is to be expected, note that the left branch of the original curve starts with negative slope, which becomes smaller until reaching eventually reaching 0 at the point where the curve changes direction and starts to increase. From that point on, its slope is positive (and getting larger). This means that the derivative of the left branch will be an increasing curve that starts negative and eventually turns positive, crossing the x-axis precisely at the point where the curve changes from decreasing to increasing.

On the other hand, the right branch of the curve has positive slope, although this slope decreases in size as we move from left to right. This means that the derivative will be positive and decreasing. Also, note how the curve begins with a perpendicular slope, which means that its derivative begins decreasing from +\infty.

Problem 3 asks us to find the derivative of \displaystyle \frac{x^3+1}{x-2}.

To do this, we use the quotient rule: \displaystyle \left(\frac fg\right)'=\frac{f'g-fg'}{g^2}. We have:

\displaystyle \frac{x^3+1}{x-2}=\frac{(3x^2)(x-2)-(x^3+1)(1)}{(x-2)^2}=\frac{2x^3-6x^2-1}{(x-2)^2}.

Problem 4 asks for the derivative of \displaystyle \sqrt{5-\sqrt x}.

Now we use the chain rule (and the power rule): \displaystyle (f(g(x)))'=f'(g(x)g'(x). In this case, we can take f(x)=\sqrt x=x^{1/2} and g(x)=5-\sqrt x=5-x^{1/2}. Then:

  • \displaystyle f'(x)=\frac 12\cdot x^{-1/2}.
  • \displaystyle g'(x)=-\frac12\cdot x^{-1/2}.
  • \displaystyle f'(g(x))=f'(5-\sqrt x)=\frac12\cdot(5-\sqrt x)^{-1/2}.

Putting this together, we see that

\displaystyle \left(\sqrt{5-\sqrt x}\right)'=\frac12\cdot(5-\sqrt x)^{-1/2}\cdot\left(-\frac12\right)x^{-1/2} \displaystyle =-\frac14(5-\sqrt x)^{-1/2}x^{-1/2}=-\frac1{4\sqrt{5x-x\sqrt x}}.

Problem 5 tells us that f(x)=(x^2+3)^2 and asks us to compute the difference quotient

\displaystyle \frac{f(x+\Delta x)-f(x)}{\Delta x}

and to simplify it as much as possible. Then, we are asked to compute the limit as \Delta x\to 0 of this expression. By definition, this limit is f'(x), so we are also asked to check that the answer we obtain coincides with the derivative as computed using the rules studied in class.

We begin with the difference quotient. For this, we start by looking at its numerator: f(x+\Delta x)-f(x).

Note that f(x+\Delta x)=((x+\Delta x)^2+3)^2.

There are two ways to proceed. We can first expand both f(x+\Delta x) and f(x) and then subtract them, or we can factor f(x+\Delta x)-f(x) as a difference of squares. The second way is less messy.

Recall that a^2-b^2=(a-b)(a+b). We have f(x+\Delta x)-f(x)=((x+\Delta x)^2+3)^2-(x^2+3)^2 =((x+\Delta x)^2+3-x^2-3)((x+\Delta x)^2+3+x^2+3) =((x+\Delta x)^2-x^2)(x^2+2 x\Delta x+(\Delta x)^2+x^2+6) =(\Delta x)(2x+\Delta x)(2x^2+2x\Delta x+(\Delta x)^2+6).

Dividing by \Delta x, we obtain

\displaystyle \frac{f(x+\Delta x)-f(x)}{\Delta x} \displaystyle =(2x+\Delta x)(2x^2+2x\Delta x+(\Delta x)^2+6).

Now, letting \Delta x\to 0, we obtain f'(x)=2x(2x^2+6).

We can verify this: Since f(x)=(x^2+3)^2, we can use the power and chain rules to compute f'(x):

f'(x)=2(x^2+3)(2x).

Clearly, this expression coincides with the one obtained by direct computation.

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This entry was posted on Sunday, September 26th, 2010 at 1:44 pm and is filed under 170: Calculus I. You can follow any responses to this entry through the RSS 2.0 feed. You can leave a response, or trackback from your own site.

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