Page 124 - Calculus Workbook For Dummies

108 Part III: Differentiation

Solutions for Derivatives and Shapes of Curves

/ 2 3
a Use the first derivative to find the local extrema of f x = 6 x - 4 x + 1. Local min at (0, 1);
^ h
local max at (1, 3).
1. Find the first derivative using the power rule.
/ 2 3
f x = 6 x - 4 x + 1
^ h
x =
f l ^ h 4 x - / 1 3 - 4
2. Find the critical numbers of f.
a. Set the derivative equal to zero and solve.
4 x - / 1 3 - 4 = 0
x - / 1 3 = 1
x = 1
4
x =
b. Determine the x-values where the derivative is undefined: f l ^ h 4 x - / 1 3 - 4 = - 4
x
3
Because the denominator is not allowed to equal zero, f l ^ h is undefined at x = 0. Thus the
x
critical numbers of f are 0 and 1.
3. Plot the critical numbers on a number line.
I’m going to skip the figure this time because I assume you can imagine a number line with
dots at 0 and 1. Don’t disappoint me!
4. Plug a number from each of the three regions into the derivative.
- / 1 3
4
1 = ^
f l ^ - h 4 - 1h - 4 = - - 4 = - 8
- / 1 3
1 1 / 1 3
f l c m = 4 c m - 4 = ^ - 4 = positive
4 2h
2 2
- / 1 3
2
f l ^ 8 = ^h 4 8h - 4 = - 4 = - 2
Note, again, how the numbers I picked for the first and third computations made the math easy.
With the second computation, you can save a little time and skip the final calculation because
all you care about is whether the result is positive or negative — this assumes that you know
that the cube root of 2 is more than 1 (you better!).
5. Draw your sign graph (see the following figure).
decreasing increasing decreasing
– + –
0 1

6. Determine whether there’s a local min or max or neither at each critical number.
f goes down to x = 0 and then up, so there’s a local min at x = 0, and f goes up to x = 1 and then
down, so there’s a local max at x = 1.
7. Figure the y-value of the two local extrema.
/ 2 3
4 0 + =
^
f 0 = ^h 6 0h - ^ h 1 1
/ 2 3
4 1 + =
f 1 = ^h 6 1h - ^ h 1 3
^
Thus, there’s a local min at (0, 1) and a local max at (1, 3). Check this answer by looking at a
graph of f on your graphing calculator.