Homework 3
1
Calculus
Homework 3
Due Date: October 24 (Wednesday)
1. Use the limit definition to find the slope of the tangent line to the graph of f at the given
point.
(a) f (x) = 4 − x2 ; (2, 0)
(b) f (x) = 6x + 3; (1, 9)
√
(c) f (x) = x + 1; (8, 3)
Solution:
(a) msec =
(b) msec =
(c) msec =
f (2+∆x)−f (2)
∆x
f (1+∆x)−f (1)
∆x
f (8+∆x)−f (8)
∆x
= −4 − ∆x. m = lim∆x→0 msec = −4.
= 6. m = lim∆x→0 msec = 6.
=
√
m = lim∆x→0 msec = 61 .
1
.
9+∆x+3
2. Use the limit definition to find the derivative of the function.
(a) f (x) = −2
(b) f (x) = 4x + 1
(c) f (x) = 4x2 − 5x
√
(d) f (x) = x + 2
(e) f (x) =
1
x+2
Solution:
f (x+∆x)−f (x)
∆x
f (x+∆x)−f (x)
lim∆x→0
∆x
f (x+∆x)−f (x)
lim∆x→0
∆x
f (x+∆x)−f (x)
lim∆x→0
∆x
(a) f 0 (x) = lim∆x→0
=0
(b) f 0 (x) =
=4
(c) f 0 (x) =
(d) f 0 (x) =
(e) f 0 (x) = lim∆x→0
f (x+∆x)−f (x)
∆x
= 8x − 5
=
√1
2 2+x
1
= − (x+2)
2
3. Find an equation of the line that is tangent to the graphs of f and parallel to the given
line.
aaaaaaFunction
Line
(a) f (x) = − 14 x2 ,
(b) f (x) = x2 − x,
Solution:
x+y =0
x + 2y − 6 = 0
2
Calculus H415611 Fall 2012
(a) f 0 (x) = − 12 x. Since the slope of given line is -1,
1
− x = −1 ⇔ x = 2 ⇒ f (2) = −1.
2
At the point (2, -1), the tangent line parallel to x + y = 0 is y = −x + 1.
(b) y = −2x − 8
4. Determine whether the statement is true or false. If it is false, explain why or give an
example that shows it is false.
(a) The slope of the graph y = x2 is different at every point on the graph of f .
(b) A tangent line to a graph can intersect the graph at more than one point.
(c) If a function is differentiable at a point, then it is continuous at that point.
(d) If a function is continuous at a point, then it is differentiable at the point.
Solution:
(a) True. The slope of the graph is given by f 0 (x) = 2x, which is different for each
different x value.
(b) True. Page 80.
(c) True. ”Differentiability implies Continuity” on page 87.
(d) False. f (x) = |x| is continuous, but not differentiable at x = 0.
5. Describe the x-values at which the function is differentiable. Explain your reasoning.
(a) y = |x + 3|
(b) y = (x − 3)2/3
(c)
x3 + 3, x < 0;
x3 − 3, x ≥ 0.
Homework 3
3
Solution:
(a) y is differentiable for all x 6= ±3. At(-3, 0) and (3, 0) the graph has a cusp.
(b) y is differentiable for all x 6= 3. At (3, 0) the graph has a cusp.
(c) y is differentiable for all x 6= 0. The function is discontinuous at x = 0.
6. Find the derivative.
(a) h(x) = 3x3
(b) f (t) = −3t2 + 2t − 4
(c) f (x) =
1
(4x)3
√
(d) f (x) = 6 x
(e) f (x) =
3
√
55x
(f) f (x) = (x2 + 2x)(x + 1)
(g) f (x) =
2x3 −4x2 +3
x2
Solution:
(a) h0 (x) = 9x2
(b) f 0 (t) = −6t + 2
(c) f 0 (x) =
(d) f 0 (x) =
(e) f 0 (x) =
−3
64x4
√3
x
3 −4/5
x
25
2
(f) f 0 (x) = 3x + 6x + 2
(g) f 0 (x) = 2 − 6x−3
7. Find an equation of the tangent line to the graph of the function at the given point.
(a) y = −2x4 + 5x2 − 3; (1, 0)
1
√
3 2
x
− x; (-1, 2)
(c) y = 3x x2 − x2 ; (2, 18)
(b) f (x) =
Solution:
(a) y 0 = −8x3 + 10x, m = y 0 (1) = 2. The equation of the tangent line at (1, 0) is
y = 2x − 2.
4
Calculus H415611 Fall 2012
(b) y =
8
x
15
+
22
.
15
(c) y = 36x − 54.
8. Determine the point(s), if any, at which the graph of the function has a horizontal tangent
line.
(a) y = −x4 + 3x2 − 1
(b) y = x2 + 2x
Solution:
√
√
(a) y 0 = −4x3 + 6x = 2x(3 − 2x2 ) = 0 when x = 0, ± 26 . If x = ± 26 , then y = 45 .
If
The
√x =0, y =−1.
function has horizontal tangent lines at the points (0, -1),
√
6 5
, , and 26 , 54 .
2 4
(b) The function has a horizontal tangent line at the point at (-1, -1).
9. Determine whether the statement is true or false. If it is false, explain why or give an
example that shows it is false.
(a) If f 0 (x) = g 0 (x), then f (x) = g(x).
(b) If f (x) = g(x) + c where c is a constant, then f 0 (x) = g 0 (x).
Solution:
(a) False. Let f (x) = x + 1 and g(x) = x. Although f 0 (x) = g 0 (x), f (x) 6= g(x).
(b) True.
10. The marginal cost for manufacturing an electrical component is $7.75 per unit, and the
fixed cost is $500. Write the cost C as a function of x, the number of units produced.
Show that the derivative of this cost function is a constant and is equal to the marginal
cost. (In economics and finance, marginal cost is the change in total cost that arises when
the quantity produced changes by one unit.)
Solution:
C = 7.75x + 500
C 0 = 7.75, which is equal to the marginal cost.