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MATH 1106, Spring 2017
Tutorial 3 Solutions
Wednesday 01/02/17
DERIVATIVES SOLUTIONS
Tutorial 3.1. From last time
Consider the function f (x) = 3x2 + 2x + 4.
(a) Find the average rate of change over the following intervals: (i)
[0, 3] (ii) [0, 2] (iii) [0, 1].
(b) Using the definition of the derivative, determine the instantaneous rate of change of f at 0.
(c) Using the definition of the derivative, determine the derivative
of a constant function F (x) = C for some real number C.
(d) * Use the definition of the derivative to show that the derivative
of xn is nxn−1 .
Hint: Try this for n = 1, n = 2 and n = 3. Can you see a pattern
emerging?
(e) Use parts (c) and (d) to find the derivative of f (x).
Solution 3.1. See solutions for Tutorial 2.
Tutorial 3.2. Tangent lines
Let y = f (x) be the graph of the function f : R → R. Recall that the
tangent line to the graph y = f (x) at x = a is defined to be the line
y − f (a) = f 0 (a)(x − a).
(a) Describe what a tangent line is. What kind of information can it
tell us about the function f ?
(b) What is the slope of the tangent line? What does it mean if the
slope is zero at x = a? Give an example of a function whose
tangent line has slope 0 at x = 0.
(c) What is the tangent line to a linear function f (x) = mx + b at any
point? How does this relate to the linear approximation interpretation of the derivative? (This question is somewhat trivial!)
(d) Let f (x) = −(x−1)2 +4. Sketch the graph of f . Find the equation
of the secant line through the points (−2, f (−2)) and (1, f (1)).
Same question for the pairs of points ((−2, f (−2)), (0, f (0))) and
((−2, f (−2)), (−1, f (−1))). Sketch these secant lines. Determine
the equation of the tangent line to the graph of f at x = −2.
Sketch this tangent line.
1
2
DERIVATIVES SOLUTIONS
(e) Let f be as in part (d). Find a point x such that the tangent line
to f is horizontal.
Solution 3.2. (a) The tangent line to the graph of a function f at the
point a is the line that is “closest” to the graph of f at a. It is the
line that best approximates f at that point. It can give an idea of
how fast the function is growing or shrinking.
(b) The slope of the tangent line is f 0 (a) (look at the equation of the
tangent line and compare it to the equation of a general line: y =
mx + b, where m is the slope), the derivative of the function f
at a. This confirms the observation from part (a). If the slope
is zero, it means that the tangent line is horizontal, given by the
equation y = f (a). Thus, at this point the function is not growing
or shrinking. An example of such a function is f (x) = x2 .
(c) Using the equation given, the equation of the tangent line is y −
f (a) = f 0 (a)(x − a) at the point a. In this case, f (a) = ma + b,
and f 0 (x) = m. Thus, the equation is y − ma − b = m(x − a).
Rearranging, we get y = ma + b + mx − ma = mx + b. The equation for the tangent line is exactly the equation of f ! This means
that the tangent line is again just the graph of f . Thinking of the
tangent line as the best linear approximation of the function f at
the point, this is exactly saying that the best linear approximation
(the “closest line”) to a line is the line itself, as we would hope!
(d) Graphs are found in Figure 1. The equation of a line is y = mx+b.
We determine m by noting that if we have two points (x1 , y1 )
and (x2 , y2 ) on the line then we have y1 = mx1 + b and y2 =
mx2 + b since the line exactly consists of the points satisfying this
equation. Thue y2 − y1 = mx2 + b − mx1 − b = m(x2 − x1 ), so
−y1
m = xy22−x
as long as x1 6= x2 . For (−2, f (−2)) and (1, f (1)), we
1)
get:
f (1) − f (−2)
4 − (−5)
m=
=
= 3.
1 − (−2)
3
Finally, since we know that the point (−2, f (−2)) is on the line, it
must satisfy the equation of the line, so:
f (−2) = −5 = m × (−2) + b = 3 × −2 + b = −6 + b
and hence rearranging, b = 1. The equation of this secant line
is y = 3x + 1. Similarly, we determine that the second secant is
given by the equation y = 4x + 3 and the third secant is given by
y = 5x + 5.
To determine the equation of the tangent line at x = −2, use
f 0 (x) = −2(x − 1) = −2x + 2, so that f 0 (−2) = 6. We also have
DERIVATIVES SOLUTIONS
3
f (−2) = −5. Hence, the equation of the tangent line is y + 5 =
6(x + 2). Rearranging, we get: y = 6x + 7.
F IGURE 1. In red: y = −(x − 1)2 + 4. Dotted lines are
the secant lines: blue is y = 3x + 1, green is y = 4x + 3,
orange is y = 5x + 5 and the purple line is the tangent
line y = 6x + 7. Note how the secants “approach” the
tangent.
(e) The tangent line is horizontal when the slope (namely the derivative) is zero. Thus we need to determine the point(s) where
f 0 (x) = 0. We saw that f 0 (x) = −2x + 2. Thus f 0 (x) = 0 when
−2x + 2 = 0. Rearranging, we get x = 1. Therefore the tangent
line to f at 1 is horizontal (check this on the graph!).
Tutorial 3.3. Definition of the derivative
(a) Using the definition of the derivative, determine the derivative
f 0 (x) of the following functions. Compute the derivative at x =
−3, x = 0 and x = 2.
(i) f (x) = x2 + 6x + 1.
(ii) f (x) = 1/x.
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DERIVATIVES SOLUTIONS
√
(iii) f (x) = 5 x.
(b) * Using the definition of the derivative, prove the rules for derivatives seen in class, namely for differentiable functions f and g
and constant c:
(i) (cf )0 = cf 0
(ii) (f + g)0 = f 0 + g 0
0
0
0
(iii) (f
·g) = f g + f g
(iv)
f
g
0
=
f 0 g−f g 0
g2
where g 6= 0.
Solution 3.3. (a) Using the definition of the derivative for each of
the following functions:
(i)
(x + h)2 + 6(x + h) + 1 − x2 − 6x − 1
h→0
h
2xh + 6h
= lim
h→0
h
h(2x + 6)
= lim
h→0
h
= lim 2x + 6
f 0 (x) = lim
h→0
= 2x + 6
Evaluating, we get f 0 (−3) = 2 · (−3) + 6 = 0, f 0 (0) = 6 and
f 0 (2) = 4 + 6 = 10.
(ii)
f 0 (x) = lim
h→0
= lim
1
x+h
−
h
1
x
x−(x+h)
x(x+h)
h
−h
= lim
h→0 h · x(x + h)
−1
= lim
h→0 x(x + h)
−1
= 2
x
h→0
Note that f (and hence its derivative) is not defined at 0.
Therefore we cannot compute f 0 (0). We have f 0 (−3) = −1
9
and f 0 (2) = −1
.
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DERIVATIVES SOLUTIONS
5
(iii) Note first that f is only defined for x positive, and hence in
this limit, we have to be somewhat careful that x + h > 0.
√
√
5 x+h−5 x
0
f (x) = lim
h→0
√
√ h
√
√
5( x + h − x)
x+h+ x
·√
= lim
√
h→0
h
x+h+ x
5(x + h − x)
= lim √
√
h→0 h( x + h +
x)
5h
= lim √
√
h→0 h( x + h +
x)
5
= lim √
√
h→0
x+h+ x
5
= √
2 x
(b) Come talk to me if you would like to help with these proofs. You
aren’t expected to do these, but an extremely important aspect of
maths is to be able to show why things work (personally, being
shown some rules without knowing why they work is somewhat
unsatisfying!).
Tutorial 3.4. Chapter 4.1 Exercise 64
The left ventricular length (viewed from the front of the heart) of a
human fetus that is at least 18 weeks old can be estimated by
l(x) = −2.318 + 0.2356x − 0.002674x2 .
where l(x) is the ventricular length (in centimeters) and x is the age
(in weeks) of the fetus.
(a) Determine a meaningful domain for this function.
(b) Find l0 (x).
(c) Find l0 (25).
Solution 3.4. (a) Since the function is estimating a length, one expectation is that the values should be positive. Using the quadratic formula, we find that l(x) = 0 at the values
√
−0.2356 + 0.23562 − 4 · 2.318 · 0.002674
x1 =
≈ 11.28
2 · (−0.002674)
and
x2 =
−0.2356 −
√
0.23562 − 4 · 2.318 · 0.002674
≈ 76.82
2 · (−0.002674)
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DERIVATIVES SOLUTIONS
l(x) is positive between these two values (can you explain why?).
Furthermore, since we’re considering the growth of a length, we
would also expect it not to decrease, so we must detrmine where
the derivative vanishes. Computing the derivative of l(x), we get
l0 (x) = 0.2356 − 2 · 0.002674x = 0.2356 − 0.005348x
by the derivative rules (you can also compute this using the formal definition of the derivative, as seen in question 3). This is
0.2356
zero when x = 0.005348
≈ 44.05. Since the minimum age is given
as 18, we see that a meaningful domain for the function is [18, 44].
(b) See part (a).
(c) l0 (25) = 0.2356 − 0.005348 · 25 = 0.1019