Practice Problems Midterm 2
Sudesh Kalyanswamy
(1) Explain why the following statement holds: It is not true, in general, that the slope of
the tangent line to x = a of f (x)g(x) is the product of the slopes of the tangent lines
of f (x) and g(x) at x = a.
(2) Suppose f (x) and g(x) are positive functions (i.e. f (x) > 0 and g(x) > 0 for all x)
and both f (x) and g(x) are increasing everywhere. Explain why f (x)g(x) is increasing
everywhere as well. Is the condition that the functions be positive necessary?
(3) Suppose f (x) and g(x) are always increasing. Explain why f (g(x)) is also increasing.
(4) Explain why a function must be both 1 − 1 and onto to have an inverse.
(5) The function f (x) = x2 , x ≥ 0 is 1 − 1. It is not onto
√ as a function f : {x ≥ 0} → R.
However, it does have an inverse, namely g(x) = x. Why doesn’t this contradict
problem (4) above?
(6) Use the chain rule to explain why the derivative of an even function is odd. (Hint:
start with f (x) = f (−x).)
(7) Suppose f (x) is everywhere differentiable, and that g(x) = 3f (2x − 1). If f 0 (3) = 1,
find g 0 (2).
(8) Suppose f (x) = (x − 2)2 , x ≥ 2.
(a) Sketch f (x).
(b) Explain why f (x) has an inverse.
(c) Sketch g(x), where g(x) is the inverse of f (x).
(d) Find an equation for g(x).
(e) Using the formula
g 0 (a) =
1
f 0 (g(a))
,
find g 0 (1).
(9) Suppose f (x) has an inverse and that g(x) = f (x − 3).
(a) How are the graphs of f (x) and g(x) related?
(b) How are the graphs of the inverse of f (x) and the inverse of g(x) related?
(10) Suppose f (x) has an inverse, call it g(x). Suppose f (−1) = 3, and that f 0 (−1) = 2.
Find the equation of the tangent line to g(x) at x = 3.
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(11) Refer to the following table in each part of this question.
x f (x) g(x)
0
1
3
1
0
1
2
2
−1
f 0 (x) g 0 (x)
1
2
2
−1
3
0
In each part below, write the equation of the tangent line to the function at the given
point.
(a) f (x)g(x) at x = 1.
(b) ef (x) at x = 2.
(c)
f (3x−2)
3g(2x−1)
at x = 1.
(d) 2g(f (x)) at x = 1.
(12) For each of the following functions, find g 0 (x) (where g(x) is the inverse of f (x)) at the
specified point without computing the inverse function explicitly. Then write down the
tangent line to g(x) at the speicified point.
(a) f (x) = x3 − 3x2 − 1, x ≥ 2 at x = −1 (Hint: f (3) = −1)
(b) f (x) = x2 , x ≤ 0, at x = 4
(c) f (x) = ex + sin(x) at x = 1
(d) f (x) = (ln(x) − 1)1/3 at x = −1 (Hint: f (1) = −1).
(e) f (x) = ln(sin(x)), 0 < x < π/2, at x = 21 ln(3) − ln(2)
(f) f (x) = x7 + 2x5 + x at x = −4
(13) Suppose f (x) is invertible and passes through the origin with slope 2. Find the tangent
line to f −1 (x) at the origin.
(14) For this problem, consider the following table, and let g(x) be the inverse of f (x) (you
may assume it exists):
x f (x) f 0 (x)
1
4
3
2
1
−1
3
5
4
4
2
0
5
3
2
(a) Find g 0 (3)
(b) Find the equation of the tangent line to y = g(x) at x = 5
(c) If we change the value of f (3) to 6, does y = f (x) still have an inverse? Explain.
(d) If we change the value of f (3) to 1, does y = f (x) still have an inverse? Explain.
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(15) Calculate the derivative of the following functions. If a point x = a is given, find the
derivative at x = a.
√
(a) f (x) = 1 + x3
(b) f (x) = cos(ex ).
(c) f (x) = ln(2x − 3)
(d) f (x) = x2 (x + 1)50 , a = 0.
√
(e) f (x) = e
2
(f) f (x) = sin2 (x2 )
1
(g) f (x) = 23x+ x
(h) f (x) = x cos(x)
(i) f (x) = x ln(x2 + 1)
2
(16) Consider f (x) = e−x . Find all points where f (x) has a horizontal tangent line.
(17) Use the product and chain rule to show that if f (x) = tan(x) =
sec2 (x) = cos12 (x) .
sin(x)
,
cos(x)
then f 0 (x) =
(18) Suppose f (x) is a differentiable function that that g(x) = (x3 −5x+2)f (x). If f (1) = 1
and f 0 (1) = −1, find g 0 (1).
Only do the following problems if Mean Value Theorem, Inflection Points, and Graphing are on the test.
(19) Consider f (x) = sin(x/2) on [π/2, 3π/2]. Explain why the Mean Value Theorem can
be applied to this function on this interval, and find c satisfying the conclusion of the
theorem.
(20) Suppose the graph in figure 1 is the graph of f 0 (x) for some function f (x).
25
20
15
10
5
-2
1
-1
-5
-10
Figure 1: Figure for problem 20
(a) What are the critical points of f (x)?
3
2
(b) Classify the critical points of f (x) as local maxima, local minima, or neither.
(c) How many inflection points does f (x) have?
(21) Sketch the graph of f (x) = x3 − 6x2 + 9x + 1. Find all local extrema, inflection points,
and mark them on the graph.
(22) State the Mean Value Theorem. Explain graphically what it means.
(23) Consider f (x) = x2 on [−2, −1]. Find c satisfying the conclusion of the Mean Value
Theorem.
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Answers:
(1) Think product rule
(2) (f (x)g(x))0 = f 0 (x)g(x) + g 0 (x)f (x) > 0 by the assumptions. Yes, it is necessary (do
you see why?).
(3) (f (g(x))0 = f 0 (g(x))g 0 (x) > 0
(4) See homework problem
(5) A function is always onto its range, so it is onto as a function to {y ≥ 0}.
(6) Even: f (x) = f (−x). Take derivatives and use chain rule on the right side to get
f 0 (x) = −f 0 (−x). So f 0 (x) is odd.
(7) 6
(8) (a) It is a shift of y = x2 two units to the right.
(b) It is 1 − 1 after restricting to x ≥ 2.
√
(c-d) It is g(x) = x + 2
(e) 1/2
(9) (a) g(x) is a shift of f (x) three units to the right.
(b) The inverse of g(x) is a shift of the inverse of f (x) three units up.
(10) y + 1 = 12 (x − 3).
(11) (a) y = 2(x − 1)
(b) y − e2 = 3e2 (x − 2)
(c) y = 2(x − 1)
(d) y − 8 = 32 ln(2)(x − 1)
(12) (a) y − 3 = 19 (x + 1)
(b) y + 2 = − 14 (x − 4)
(c) y = 12 (x − 1)
(d) y − 1 = 3(x + 1)
√
(e) y − π3 = 3(x − 12 ln(3) + ln(2))
(f) y + 1 =
1
(x
18
+ 4)
(13) y = 12 x
(14) (a) 1/2
(b) y − 3 = 14 (x − 5)
(c) Yes
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(d) No
(15) (a)
2
√3x
2 1+x3
x
(b) −e sin(ex )
(c)
2
2x−3
(d) 0
(e) 0
(f) 4x sin(x2 ) cos(x2 )
1
(g) 3 − x12 23x+ x ln(2)
(h) −x sin(x) + cos(x)
(i)
2x2
x2 +1
+ ln(x2 + 1)
(16) x = 0.
(17) Write f (x) = sin(x)(cos(x))−1 and use product rule on this.
(18) 0
(19) c = π.
(20) x = −2 is a local maxima, x = 0 is a local minima, and x = 2 is neither. There are
three inflection points of f (x).
(21) Consult graphing calculator or WolframAlpha
(22) Consult textbook.
√
(23) c = − 3.
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