A turning point occurs when the velocity is equal to zero, but the acceleration is not equal to zero.
t(x)=(x+5)^3+7
dt/dx=3(x+5)^2
d2t/dx2=6(x-5)
dt/dx=0 only when x=-5
However, since d2t/dx2(-5)=0, this point is an inflection point, not a turning point.
So there is no turning point for this function.
Now in this problem, it is even easier than the above to show that there is no turning point. A turning point by definition is when the derivative or velocity changes sign. Since in this case v=3(x+5)^2, for any value of x, v≥0, and thus never becomes negative, so it never changes from a positive to negative velocity because velocity in this instance is a squared function.
Answer:
2 ≤ x
Step-by-step explanation:
The perpendicular line will have the x and y coefficients swapped and one of them negated. We can write the desired line as
9(x -6) -3(y -4) = 0
where the coordinates of the point of interest are (6, 4).
Dividing by 3, this is
3(x -6) -(y -4) = 0
3x -y = 14
An equation for the line of interest is ...
3x -y = 14
Answer:
45 mph
Step-by-step explanation:
This is a really good question to know the answer to. It is tricky and a bit indirect (which means you have to find something else before you can find the speed of the car.)
Let's keep track of what he does in the time allotted.
How far does Joe go in 5 minutes? That's the amount of time he's on the road before she is.
convert 5 minutes into hours. 5 minutes * 1 hour / 60 minutes = 1/12 of an hour
d = r*t
r = 30 km/hour
t = 1/12 hour
d = 30 km/hr * 1/12 hour = 2.5 km
Now she's about to start. She wants to catch him in 10 minutes
d = r*t
r = x mph
t = 10 minutes = 10 minutes * 1 hour * 60 minutes = 1/6 of an hour.
How far does he go in the 10 minute time?
d = 30 * 1/6 = 5 km
What is his total distance
5 km + 2.5 km = 7.5 km
Finally how fast does she need to go to catch him
d = 7.5 km
r = ? This is what you are trying to find
t = 1/6 of an hour
d = r*t
7.5 km = r * (1/6)hour divide by 1/6 hour
7.5 km // 1/6 hour = r
r = 7.5 * 6 = 45 mph
