Math 1151 Midterm 2 2021
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Math 1151 Midterm 2
February 15, 2021
1) The graph of a function f with domain (-6, 6) is given in the igure below.
a) (1 point) f (-2) = c) (2 points) 
=
b) (1 point) f' (-1) =
d) (2 points) Sketch the secant line through the points corresponding to x = 1 and x = 4. Label it as S.
e) (2 points) Sketch the line tangent to f at the point corresponding to x = -3. Label it as T.
f) (2 points) Which one of these values is greatest? A, B, or C :
(A) f' (-5.5) (B) f' (-2.25) (C) f' (4.5)
g) (2 points) Which one of these values is least? A, B, or C:
(A) f' (-5.5) (B) f' (-2.25) (C) f' (4.5)
h) (3 points) Find the x-values in (-6, 6) at which f is not diferentiable?
x - VALUES:
2) We are given that the line y = 3x - 7 is tangent to the graph of y = f (x) at the point (2, f (2)) (and only at that point). Set g (x) = 2xf (
).
a) (2 points) What is the value of f (2)?
f (2) =
b) (2 points) What is the value of f、(2)?
f' (2) =
c) (2 points) What is the value of g (4)?
g (4) =
d) (5 points) What is the value of g、(4)?
g' (4) =
e) (4 points) Find an equation for the line tangent to the graph of y = g (x) at x = 4.
EQUATION :
3) A function g , with domain (-1, 1), has values and derivative values as given in the table below.
Evaluate the following. Don’t forget to JUSTIFY your work.
a) (5 points) 
VALUE :
b) (5 points) 
VALUE :
c) (5 points) 
VALUE :
4) An object traveling along a horizontal line has displacement function given by
s(t) = 4t(t + 1)
with s measured in meters and t measured in minutes.
a) (4 points) Find the instantaneous velocity at time t = 1, v(1).
v(1) =
b) (4 points) Find a formula for the average velocity, vav (t), on the time interval [t, 1] for 0 < t < 1, and on [1, t] for 1 < t < 2.
vav (t) =
c) (4 points) Using your result from b), compute the limit: 
VALUE :
d) (3 points) Explain what the limit from part c) represents.
2024-03-29