Exercise 13.1
1
2
a) y = sin x − cos x ⇒
dy
= cos x − (− sin x ) = cos x + sin x
dx
b) y = 5 − e x ⇒ dy = 0 − e x = − e x
dx
c) y = x + ln x ⇒ dy = 1 + 1  or x + 1
dx
x 
x 
x
2
2e x
2e
dy 2
d) y =
= × e2 ⇒
= × ex =
5
5
dx 5
5
dy
e) y = x 3 + 2 cos x ⇒
= 3x 2 + 2 (− sin x ) = 3x 2 − 2 sin x
dx
f) y = 2e ln x ⇒ dy = 2e × 1 = 2e
dx
x
x
 π 3
3
1
π
π
π
π
a) x = ⇒ y = sin   =
⇒ P ,
y ' = cos x ⇒ m = y '   = cos   =







3
3
2
3
3
2
3 2 
Equation of tangent line: y =
π
3
1
3 3−π
1
3 π 1
x− +
⇒y= x+
− = x+
2
3
2
2
2
6 2
6
b) x = 0 ⇒ y = 0 + e 0 = 1 ⇒ P (0, 1)
y ' = 1 + e x ⇒ m = y ' (0) = 1 + e 0 = 2
Equation of tangent line: y = 2 ( x − 0) + 1 ⇒ y = 2 x + 1
1
1
1 1
1
1
1
c) x = e ⇒ y = ln e = ⇒ P  e , 
y'= × =
⇒ m = y ' (e) =
 2
2
2
2 x 2x
2e
1
1
1
1
1
1 1
1
1
(x − e) + ⇒ y = x −
×e + ⇒y=
x− + ⇒ y=
x
2e
2
2e
2e
2
2e
2 2
2e
5π
1
π
a) y ' = 1 + 2 (− sin x ) = 1 − 2 sin x ⇒ y ' = 0 ⇒ 1 − 2 sin x = 0 ⇒ sin x = ⇒ x = or x =
6
6
2
Equation of tangent line: y =
3
4
5
6

π
π
3
 y "   = −2 cos   = −2 ×
= − 3 < 0 ⇒ maximum




6
6
2

b) y " = 0 − 2 cos x = −2 cos x ⇒ 
 y "  5π  = −2 cos  5π  = −2 ×  − 3  = 3 > 0 ⇒ minimum
 2 
 6
 6


y ' = 1 − e x ⇒ y ' = 0 ⇒ 1 − e x = 0 ⇒ e x = 1 ⇒ x = 0 ⇒ y = 0 + e 0 = 1 ⇒ P (0, 1)
y " = 0 − e x = −e x ⇒ y " (0) = −e 0 = −1 < 0 ⇒ P (0, 1) is a maximum point.
1
1
1
y ' = 1 − = 1 − x −1 ⇒ y " = 0 − ( −1) x −2 = 2 ⇒ y " ( x ) = 0 ⇒ 2 = 0 ⇒ x ∉ .
x
x
x
Therefore, there are no points of inflexion.
π
π
π
π
y = 3 + sin   = 3 + 1 = 4 ⇒ P  , 4
y ' = cos x ⇒ mT = y '   = cos   = 0 ⇒ mN is not
 2
2 
 2
 2
defined. Therefore, the normal line is a vertical line through point P and its equation is x =
7
a) f ' ( x ) = e x − 3x 2 ⇒ f " ( x ) = e x − 6 x
π
.
2
Exercise 13.1
b)
c) T
o answer this part, we are going to refer to the graph of the derivative function given in part b when
zeros of the derivative function are found.
The function f increases whenever f ' ( x ) > 0 ⇒ x ∈ ]−0.459, 0.910[ ∪ ]3.73, ∞[ .
The function f decreases whenever f ' ( x ) < 0 ⇒ x ∈ ]−∞ , −0.459[ ∪ ]0.910, 3.73[ .
d) A
t x = −0.459 and x = 3.73 we have minimum points since the derivative function changes its sign
from negative to positive and therefore the original function changes from decreasing to increasing,
which is a minimum point. At x = 0.910 we have a maximum point since the derivative changes
its sign from positive to negative and therefore the original function changes from increasing to
decreasing, which is a maximum point.
e)
f) T
o answer this part, we are going to refer to the graph of the second derivative function given in
part e when zeros of the second derivative function are found.
The function f is concave down whenever f " ( x ) < 0 ⇒ x ∈ ]0.204 , 2.83[ .
1
1
1
1
y ' = ⇒ m = y ' (x0 ) =
⇒T : y =
(x − x 0 ) + y0 = x − 1 + y0 must be a line that passes
x
x0
x0
x0
through the origin; therefore, the y-intercept must be equal to zero. Hence,
1
−1 + y0 = 0 ⇒ y0 = 1 ⇒ ln x 0 = 1 ⇒ x 0 = e ⇒ m = .
e
We can confirm the result by drawing
the function and the tangent using a GDC.
Notice that the coordinates of the touching point are
not exactly (e, 1), but they are fairly close to it. The
numerical algorithm that the calculator uses introduces a certain accuracy error.
1
1
1
1
ln x
y = log b x =
=
× ln x ⇒ y ' =
× =
ln b ln b
ln b x x ln b
8
9
There are three possible answers: x = −0.459, x = 0.910, or x = 3.73 .
There are two possible points of inflexion: x = 0.204 or x = 2.83 .
The function f is concave up whenever f " ( x ) > 0 ⇒ x ∈ ]−∞ , 0.204[ ∪ ]2.83, ∞[.