Is there another side to it?
Answer:
a
Step-by-step explanation Option 1:
Plug
4
into
g
(
x
)
:
g
(
x
)
=
−
2
(
4
)
−
6
=
−
8
−
6
=
−
14
Then plug
g
(
x
)
into
f
(
x
)
:
f
(
x
)
=
3
(
−
14
)
−
7
=
−
42
−
7
=
−
49
Option 2:
Plug
g
(
x
)
into
f
(
x
)
:
f
(
x
)
=
3
(
−
2
x
−
6
)
−
7
=
−
6
x
−
18
−
7
=
−
6
x
−
25
Finally plug
4
into our current
f
(
x
)
:
f
(
x
)
=
−
6
(
4
)
−
25
=
−
24
−
25
=
−
49
Answer:
4/10= 0.4
3 1/10= 3.1
7/10= 0.7
6 5/10= 6.5
9/10= 0.9
Step-by-step explanation:
Answer:
5cm
Step-by-step explanation:
40/8=5
1*5=5
The distance walked will be (12-x) km
the distance rowed will be (9+x²) km
A] The function T(x) will be given by:
Time=distance/speed
thus we shall have:
T(x)=[√(9+x²)]/2.5+(12-x)/4
B] To get the distance x=c that minimizes the time travel, we differentiate the above.
T'(x)=(1/2.5)[1/(2.5√9+x²)*2x-1/4]
this should give us 0 for x=c, thus
c/[2.5*√(9+c²)]-1/4=0
⇒c/[2.5*√(9+c²)]=1/4
c/√(9+c²)=2.5/4
squaring both sides we get:
c²/(9+c²)=5/8
8c²=5(9+c²)
8c²=45+5c²
3c²=45
c²=15
c=3.87 km
c] The least travel time is
T(c)=[√(9+c²)]/2.5+(12-c)/4
this will give us:
T(c)=[√(9+3.87²)]/2.5+(12-3.87)/4
T(c)=3.9999209~4 hours
d] The second derivative will be:
T"(x)=1/[2.5√(9+x²)]-x²/[2.5(9+x²)^(3/2)]
but
x=c=3.87
T"(x)=0.01668 hours/ mile²
Given that T(c)=0, while T(x)<0 for x<c and T(x)>0 for x>c proves that T(x) decreases for x<c and increases for x>c, so there is a minimum at x=c
