Department of Mathematical Sciences
Instructor: Daiva Pucinskaite
Calculus I
June 8, 2016
Quiz 5
Consider the following function f given by
f (x) = x3 − 4x2 + 2x − 1
a. Find the derivative of f
Recall: Let f and g be some functions.
• The Sum rule: (f (x) ± g(x))0 = f 0 (x) ± g 0 (x) (other notation
0
0
• The Constant Multiple Rule: (cf (x)) = cf (x) (other
constant
d
(f (x) + g(x))
dx
d
(cf (x))
notation dx
=
=
d
d
f (x) + dx
g(x))
dx
d
c dx f (x)) where c
is a
• For f (x) = xn we have f 0 (x) = (xn )0 = nxn−1 , n is an integer.
f 0 (x)
=
=
constant rule
=
=
=
sum rule
(x3 − 4x2 + 2x − 1)0
(x3 )0 − (4x2 )0 + (2x)0 − (1)0
(x3 )0 − 4(x2 )0 + 2(x)0 − (1)0
3x2 − 4 · 2x + 2 − 0
3x2 − 8x + 2
b. Which of the following points lies on the graph of the function f .
(−1, −8),
(0, 0),
(−1, 8).
Recall: A point (a, b) is on the graph of a function f , if f (a) = b.
• Since f (−1) = (−1)3 − 4(−1)2 + 2(−1) − 1 = −1 − 4 − 2 − 1 = −8, the point (−1, −8)
lies on the graph of the function f given by f (x) = x3 − 4x2 + 2x − 1.
• Since f (0) = 03 − 4 · 02 + 2 · 0 − 1 = −1 6= 0, the point (0, 0) lies not on the graph of
the function f given by f (x) = x3 − 4x2 + 2x − 1.
• Since f (−1) = (−1)3 − 4(−1)2 + 2(−1) − 1 = −8 6= 8, the point (−1, 8) lies not on
the graph of the function f given by f (x) = x3 − 4x2 + 2x − 1.
c. Find the slope of the tangent line to the graph of f at the point determined in b.
Recall: Let f be a function, and we can draw a tangent line at the point (a, f (a)), then


the slope of the tangent line
 to the graph of the function f  = f 0 (a) = lim f (a + h) − f (a)
h→0
h
at the point (a, f (a))
Since f 0 (x) = 3x2 − 8x + 2,

the slope of the tangent line
 to the graph of the function f


at the point (−1, f (−1))
| {z }
−8


 = f 0 (−1) = 3(−1)2 − 8(−1) + 2 = 13.
