Download Fall 2016 Math 19 - HW 4 solutions Problem 1 (1 review #12). For f(n

Fall 2016 Math 19 - HW 4 solutions
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Problem 1 (1 review #12). For f (n) = 3n2 − 2 and g(n) = n + 1, find and simplify:
(a) f (n) + g(n)
(b) f (n)g(n)
(c) The domain of f (n)/g(n)
(d) f (g(n))
(e) g(f (n))
Solution.
(a) f (n) + g(n) = 3n2 − 2 + n + 1 = 3n2 + n − 1
(b) f (n)g(n) = (3n2 − 2)(n + 1) = 3n3 + 3n2 − 2n − 2.
(c) By definition, n belongs to the domain of the function f /g if and only if n belongs
to the domain of both f and g and g(n) 6= 0. Since we know that f and g are
well-defined on the set of all real numbers, we just need to find points t such that
g(t) = 0. This means that t + 1 = 0, so t = −1. Thus, we conclude that the domain
of f /g is R \ {−1} (all real numbers not equal to −1).
(d) f (g(n)) = f (n + 1) = 3(n + 1)2 − 2 = 3(n2 + 2n + 1) − 2 = 3n2 + 6n + 1.
(e) g(f (n)) = g(3n2 − 2) = (3n2 − 2) + 1 = 3n2 − 1.
Problem 2 (1 review #16). Solve the following equation for t using logs.
7 · 3t = 5 · 2t
Solution.
7 · 3t
7 · 3t
5 · 3t
7
5
7
5
= 5 · 2t
5 · 2t
=
5 · 3t
2t
= t
3
t
2
=
3
Now, in order to get the solution we need to apply log2/3 to both sides of the equation.
t = log2/3 (7/5)
Other equivalent answers:
t=
ln(7/5)
ln 7 − ln 5
ln 5 − ln 7
=
=
= log3/2 (5/7)
ln(2/3)
ln 2 − ln 3
ln 3 − ln 2
1
Problem 3 (1 review #48). The graph of Fahrenheit temperature ◦ F, as a function of
Celsius temperature, ◦ C, is a line. You know that 212 ◦ F and 100 ◦ C both represent the
temperature at which water boils. Similarly, 32 ◦ F and 0 ◦ C both represent water freezing
point.
(a) What is the slope of the graph.
(b) What is the equation of the line.
(c) Use the equation of find what Fahrenheit temperature corresponds to 20 ◦ C.
(d) What temperature is the same number of degrees in both Celsius and Fahrenheit?
Solution. (a) The slope is equal to the quotient of the differences in degrees at two
= 180
= 1.8 = 59 .
points. Thus, the slope is given by the formula 212−32
100−0
100
(b) The problem actually asks us to find a line that passes through the points (212, 100)
and (32, 0). We have already proven that that the line is of the form y = 1.8x + b.
So, we just need to plug into this equation coordinates (x, y) of any point and find
b (it does not matter which point, since we know that the line exists and is unique).
Plugging x = 32 and y = 0 we obtain b = 32. Therefore the equation we are looking
for is y = 1.8x + 32.
(c) Just plug 20 in x in the equation of the graph to get Fahrenheit temperature corresponding to 20 ◦ C. That is, y = 1.8 × 20 + 32 = 68, so the answer is 68.
(d) To find such a temperature we need to find a point on the graph such that its xcoordinate is the same as y-coordinate. This is equivalent to solving of the following
equation:
x = 1.8x + 32
−0.8x = 32
x = −40.
Thus the answer is −40 ◦ C (or ◦ F, which is in the case the same).
Problem 4 (1 review #64). The half-life of radioactive strontium-90 is 29 years. In
1960, radioactive strontium-90 was released into the atmosphere during testing of nuclear
weapons, and was absorbed into people’s bones. How many years does it take until only
10% of the original amount absorbed remains?
Solution. Denote the number of years we are looking for by t. Then the problem says us
that for every 29 years the amount of absorbed strontium-90 reduces by 2 times. Thus
we have an equation for t:
0.1 = (1/2)t/29 .
Apply log1/2 to both sides of the equation:
log1/2 (1/10) = t/29
2
− log1/2 (10) = t/29
t = −29 log1/2 (10)
In order to simplify this solution a little bit more we need to use the following formula
log1/a (b) = − loga (b) to obtain the final answer t = −29 log1/2 (10) = 29 log2 (10) ≈ 96.336
years.
Problem 5 (2.1 #18). Match the points labeled on the curve in Figure 2.8 with the given
slopes (-3, -1, 0, 1/2, 1, 2)
Solution. Let us denote the slope of the graph at the point x by sl(x). Note that the
point E is a vertex of this graph, so the slope of this point is 0. Also, the graph tells us
that the function is increasing at the points A, B and D, so the slopes at these points
are positive numbers. Since the graph grows faster at the point D than at the point B,
and at the point B it grows faster than at the point A, we conclude sl(D) > sl(B) >
sl(A) > 0.Therefore, D stands for 2, B stands for 1 and A stands for 1/2. The same
comparison of the behaviour of the graph at points C and F shows that the absolute
value of the slope at the point F is greater than at the point C. Hence, the answer is
(−3, D), (−1, C), (0, E), (1/2, A), (1, B), (2, D)
Problem 6 (2.1 #20). . For the graph in Figure 2.10, list the following quantities in
order, smallest to largest.
(a) The slope of the graph at A
(b) The slope of the graph at B
(c) The slope of the graph at C
(d) The slope of the line AB
(e) The number 0
(f) The number 1
Solution. First note that all these quantities are greater than or equal to 0, meaning 0 is
the smallest. It’s also important to note that 1 is the slope of the line y = x, which is
drawn on the graph. Comparing steepness, we see that the tangent line of the graph at
A is steeper than the line y = x. The line AB is not as steep as the line y = x, but is
steeper than the tangent line of the graph at B. The tangent line of the graph at C is
the flattest. That gives the following order:
0 < slope at C < slope at B < slope of AB < 1 < slope at A
3
(1 + h)3 − 1
h→0
h
Problem 7 (2.1 #26). lim
Solution. Carefully expanding, we find that (1 + h)3 = 1 + 3h + 3h2 + h3 . Then, we have
that
(1 + h)3 − 1
1 + 3h + 3h2 + h3 − 1
= lim
h→0
h→0
h
h
3h + 3h2 + h3
= lim
h→0
h
h(3 + 3h + h2 )
= lim
h→0
h
= lim (3 + 3h + h2 ) = 3.
lim
h→0
Alternate solution. The idea is to use the following formula x3 − y 3 = (x − y)(x2 + xy + y 2 )
of abridged multiplication to simplify the expression. Namely, we have
(1 + h)3 − 13
(1 + h)3 − 1
= lim
lim
h→0
h→0
h
h
(1 + h − 1)((1 + h)2 + (1 + h) + 1)
= lim
h→0
h
2
h(3 + 3h + h )
= lim
h→0
h
= lim (3 + 3h + h2 ) = 3.
h→0
(3 + h)2 − (3 − h)2
h→0
2h
Problem 8 (2.1 #28). lim
Solution.
9 + 6h + h2 − (9 − 6h + h2 )
(3 + h)2 − (3 − h)2
= lim
h→0
h→0
2h
2h
12h
= lim
= lim 6 = 6
h→0 2h
h→0
lim
Alternate solution. The idea is the same as in the previous exercise. The only difference
is that we need to use another formula of abridged multiplication. Namely, we are going
to use the following one: x2 − y 2 = (x − y)(x + y). We can simplify the expression
(3 + h)2 − (3 − h)2
((3 + h) − (3 − h))((3 + h) + (3 − h))
lim
= lim
=
h→0
h→0
2h
2h
(2h)(6)
= lim
= lim 6 = 6.
h→0
h→0
2h
4
√
1/ 4+h−1/2
.
h
√
expression 1/ 4+h−1/2
h
Problem 9 (2.2 #40). Evaluate the limit limh→0
√
1
√ h+4 .
Solution. First of all, let us simplify this
= h√1h+4 − 2h
= 2−
2h h+4
Now in order to compute it we will multiply the numerator and denominator by the
conjugate of the numerator.
√
√
2− h+4
1/ 4 + h − 1/2
√
= lim
lim
h→0 2h h + 4
h→0
h
√
√
(2 − h + 4)(2 + h + 4)
√
√
= lim
h→0 2h h + 4(2 +
h + 4)
4 − (h + 4)
√
= lim √
h→0 2h h + 4(2 +
h + 4)
−h
√
= lim √
h→0 2h h + 4(2 +
h + 4)
−1
√
= lim √
h→0 2 h + 4(2 +
h + 4)
1
−1
√ =−
= √
16
2 4(2 + 4)
Problem 10 (2.2 #50). Find the line tangent to the graph of the function f (x) = 1/x2
at (1, 1).
Solution. We know that for any continuously differentiable function f the line tangent
to any point (x0 , y0 ) of its graph is of the form y = f 0 (x0 )x + b, where b is uniquely
determined by the equation y0 = f 0 (x0 )x0 + b. Let’s apply it for (x0 , y0 ) = (1, 1) and
f (x) = 1/x2 . In this case,
0
f (x) = lim
h→0
1
(x+h)2
h
−
1
x2
1
x2
(x + h)2
= lim
−
h→0 h
x2 (x + h)2 x2 (x + h)2
1 x2 − (x2 + 2xh + h2 )
= lim
h→0 h
x2 (x + h)2
1 −2xh − h2
= lim
h→0 h
x2 (x + h)2
−2x − h
= lim 2
h→0 x (x + h)2
2x
2
=− 4 =− 3
x
x
Thus, f 0 (1) = −2. And we also need to solve the equation 1 = −2 · 1 + b in order to find
b. One has b = 1 + 2 = 3. Hence the line tangent to the graph y = 1/x2 at the point (1, 1)
is given by the equation y = −2x + 3.
5
Problem 11 (2.3 #30). Sketch the graph of f 0 (x).
Solution. First of all, note that the graph of f is line that passes through points (0, 2)
and (1, 0). Hence, we know that its derivative is constant and is equal to the slope of this
line. Let us compute it, it is equal to ∆(y)/∆(x) = (f (1) − f (0))/1 = −2. Thus, the slope
equals to −2 and the graph of f 0 is a horizontal line y = −2.
3
2
1
x
−5
−4
−3
−2
−1
1
2
3
4
5
−1
−2
−3
Problem 12 (2.3 #32). Sketch the graph of f 0 (x).
Solution. Note that the graph of f is parabola. Therefore, f is a quadratic function. This
implies that the function f 0 is linear. Now, according to the picture the vertex of this
parabola has x coordinate equal to 1. This means that that the derivative of f vanishes
at this point. So, we know that the graph of f 0 is a line passes through the point (1, 0).
Also, note that f is a monotonically decreasing function on the interval (1, +∞). So, its
derivative f 0 (x) is negative for x > 1. The same argument shows that f 0 (x) is positive for
x < 1. Based on these observations, we conclude that the graph of f 0 is of the following
form:
3
2
1
x
−5
−4
−3
−2
−1
1
−1
−2
−3
6
2
3
4
5