CHAPTER 8
8.40
8.41
8.42
8.44
Fig. 8-1
Fig. 8-2
Figure 8-2 shows the graph of the function f(x) = x2 -4x. Draw the graph of y = \f(x)\ and determine
where y' does not exist.
Fig. 8-3
54
8.38
Find the point(s) on the graph of y = x
2
at which the tangent line passes through (2, —12).
The slope of the tangent line is the derivative 2x. Since (x, x
2 ) and (2, -12) lie on the tangent line, its
slope is (x2 + 12) l(x - 2). Hence, (x2 + 12) /(x - 2) = 2x, X2 + 12 = 2x2 -4x, x2 - 4* - 12 = 0, (x - 6)(x +
2) = 0, x = 6 or x = -2. Thus, the two points are (6,36) and (-2,4).
8.39
Use the A-defmition to calculate the derivative of f(x) = x
4
.
Find a formula for the derivative D x [f(x) g(x) h(x)].
Find
By Problem 8.40, x(2x - 1) + x • 2 • (x + 2) + (2* - l)(x + 2) = x(2x -1) + 2x(x + 2) + (2* - 1)(* + 2).
Let /(*) = 3x
3 — llx
2 — 15x + 63. Find all points on the graph of/where the tangent line is horizontal.
The slope of the tangent line is the derivative f'(x) = 9x -22*-15. The tangent line is horizontal when
and only when its slope is 0. Hence, we set 9x
2 - 22x - 15 = 0, (9* + 5)(* - 3) = 0, x-3 or *=-!.
Thus, the desired points are (3,0) and
8.43
Determine the points at which the function f(x) = \x - 3| is differentiable.
The graph (Fig. 8-1), reveals a sharp point at x = 3, y — 0, where there is no unique tangent line. Thus
the function is not differentiable at x = 3. (This can be verified in a more rigorous way by considering the
A-definition.)
Hence
Dx[x(2x-1)()(x+2)].
So,
By the product rule, £>,{[/(*) g(x)]h(x)} = /(*) g(x) h'(x) + D x [f(x) g(x)]h(x) = f(x)g(x)h'(x) +
[fWg'(x)+f'(x)g( X)]h(X)=f(X)g(X)h'(X)+f(X)g'(X)h(X)+f'(X) g(X) h(X).