Name:
MATH 151:513 - Quiz 4
Instructions. Time: 10 minutes. Please, no notes, no calculator, no cellphone, no headphones, no
electronic devices. Fill the blanks to answer the questions.
This exercise will guide you through the computation of the tangent line to the graph of a
given function f (x) at a given point x = x0 .
Exercise 1.
Consider f (x) = 2x3 + 1 and x0 = −2.
(2pt) Compute f (x0 ) and f (x0 + h) for a generic h. If possible, expand and simplify f (x0 + h).
f (x0 ) = 2(−2)3 + 1 = −15,
f (x0 + h) = 2(−2 + h)3 + 1 = 2(−8 + 3 · 4h + 3(−2)h2 + h3 ) + 1 = −15 + 24h − 12h2 + 2h3 .
(2pt) Compute msec =
f (x0 + h) − f (x0 )
. Simplify the expression.
h
f (x0 + h) − f (x0 )
(−15 + 24h − 12h2 + 2h3 ) + 15
=
=
h
h
=
h(24 − 12h + 2h2
= 24 − 12h + 2h2
h
(2pt) Compute the limit mtan = lim msec .
h→0
mtan = lim msec = lim (24 − 12h + 2h2 ) = 24
h→0
h→0
(2pt) Compute the equation of the tangent line, in the form y = mx + b.
btan = f (x0 ) − mtan x0 = −15 − 24(−2) = 33
y = 24x + 33
(2pt) Can you explain briey the geometric meaning of msec and mtan ?
msec is the slope of the secant line to the graph of f (x) passing through (x0 , f (x0 )) and
(x0 + h, f (x0 + h));
mtan is the slope of the tangent line to the graph of f (x) at the point (x0 , f (x0 )).
Name:
MATH 151:514 - Quiz 4
Instructions. Time: 10 minutes. Please, no notes, no calculator, no cellphone, no headphones, no
electronic devices. Fill the blanks to answer the questions.
This exercise will guide you through the computation of the tangent line to the graph of a
given function f (x) at a given point x = x0 .
Exercise 1.
Consider f (x) = 3x3 + 1 and x0 = −1.
(2pt) Compute f (x0 ) and f (x0 + h) for a generic h. If possible, expand and simplify f (x0 + h).
f (x0 ) = 3(−1)3 + 1 = −2,
f (x0 + h) = 3(−1 + h)3 + 1 = 3(−1 + 3h − 3h2 + h3 ) + 1 = −2 + 9h − 9h2 + 3h3 .
(2pt) Compute msec =
f (x0 + h) − f (x0 )
. Simplify the expression.
h
f (x0 + h) − f (x0 )
(−2 + 9h − 9h2 + 3h3 ) + 2
=
=
h
h
=
h(9 − 9h + 3h2
= 9 − 9h + 3h2
h
(2pt) Compute the limit mtan = lim msec .
h→0
mtan = lim msec = lim (9 − 9h + 3h2 ) = 9
h→0
h→0
(2pt) Compute the equation of the tangent line, in the form y = mx + b.
btan = f (x0 ) − mtan x0 = −2 − 9(−1) = 7
y = 9x + 7
(2pt) Can you explain briey the geometric meaning of msec and mtan ?
msec is the slope of the secant line to the graph of f (x) passing through (x0 , f (x0 )) and
(x0 + h, f (x0 + h));
mtan is the slope of the tangent line to the graph of f (x) at the point (x0 , f (x0 )).
Name:
MATH 151:515 - Quiz 4
Instructions. Time: 10 minutes. Please, no notes, no calculator, no cellphone, no headphones, no
electronic devices. Fill the blanks to answer the questions.
This exercise will guide you through the computation of the tangent line to the graph of a
given function f (x) at a given point x = x0 .
Exercise 1.
Consider f (x) = −x3 + 3 and x0 = 3.
(2pt) Compute f (x0 ) and f (x0 + h) for a generic h. If possible, expand and simplify f (x0 + h).
f (x0 ) = −33 + 3 = −24,
f (x0 + h) = −(3 + h)3 + 3 = −(27 + 3 · 9h + 3 · 3h2 + h3 ) + 3 = −24 − 27h − 9h2 − h3 .
(2pt) Compute msec =
f (x0 + h) − f (x0 )
. Simplify the expression.
h
f (x0 + h) − f (x0 )
(−24 − 27h − 9h2 − h3 ) + 24
=
=
h
h
=
h(−27 − 9h − h2 )
= −27 − 9h − h2 .
h
(2pt) Compute the limit mtan = lim msec .
h→0
mtan = lim msec = lim (−27 − 9h − h2 ) = −27
h→0
h→0
(2pt) Compute the equation of the tangent line, in the form y = mx + b.
btan = f (x0 ) − mtan x0 = −24 − (−27)3 = 57,
y = −27x + 54.
(2pt) Can you explain briey the geometric meaning of msec and mtan ?