So I'm assuming that you're taking Calculus.
The first thing you want to do is take the integral of f(x)...
Use the power rule to get:
4X^2-13X+3.
Now solve for X when f(x)=0. This is because when the slope is 0, it is either a minimum or a maximum(I'm assuming you know this)
Now you get X=0.25 and X=3. Since we are working in the interval of (1,4), we can ignore 0.25
Thus our potential X values for max and min are X=1,X=4,X=3(You don't want to forget the ends of the bounds!)
Plugging these value in for f(x), we get
f(1)=2.833
f(3)=-8.5
f(4)=1.667
Thus X=1 is the max and X=3 is the min.
So max:(1,2.833)
min:(3,-8.5)
Hope this helps!
Answer:
c - 7 < -18
Step-by-step explanation:
Seven subtracted from c : c - 7
c - 7 < -18
Answer:
b(b/a)^2
Step-by-step explanation:
Given that the value of the car depreciates such that its value at the end of each year is p % less than its value at the end of the previous year and that car was worth a dollars on December 31, 2010 and was worth b dollars on December 31, 2011, then
b = a - (p% × a) = a(1-p%)
b/a = 1 - p%
p% = 1 - b/a = (a-b)/a
Let the worth of the car on December 31, 2012 be c
then
c = b - (b × p%) = b(1-p%)
Let the worth of the car on December 31, 2013 be d
then
d = c - (c × p%)
d = c(1-p%)
d = b(1-p%)(1-p%)
d = b(1-p%)^2
d = b(1- (a-b)/a)^2
d = b((a-a+b)/a)^2
d = b(b/a)^2 = b^3/a^2
The car's worth on December 31, 2013 = b(b/a)^2 = b^3/a^2
Answer:
6x+2y+3=0
Step-by-step explanation:
y=-3x-(3/2) add 3x
y+3x=-(3/2) divide -1
-y-3x=3/2 multiply 2
-2y-6x=3 subtract 3
-2y-6x-3=0 multiply -1
2y+6x+3=0 rearrange
Answer:20% OF THEM WERE SHIPPED INTERNATIONALLY IT IS THE BLUE MODEL NOT THE PINK ONE
Step-by-step explanation:
