At what x-coordinate does f(x) = |5x + 1| have its minimum value?

A) -1
B) -1/5
C) 1
D) 1/5

B.

Step-by-step explanation:

2 ways to solve this: using calculus or not using calculus

not using calculus: we know |x| + b will move graph up by b amount of units, but since there is no b, the minimum f(x) value is f(x) = 0

finding x now

0 = |5x+1|

since 0 is neither positive nor negative, we can remove absolute value signs without creating 2 equations

0 = 5x + 1

- 1 = 5x

x =

using calculus (disregard if you are not in calculus)

setting f(x) = y

finding critical points, where is 0 or undefined

is undefined at x =

testing left and right sides of our critical points to see if it is a max or min point

use easy values for this

at x = 0, = 5

at x = -1, = -5

since goes from negative to positive at x = ,

x = is a relative minimum

since x = is our only critical point, it is also our absolute minimum (relative minimums do not always correspond to absolute minimums)

this is false ^.^ lol

well we know there are 4 clubs in 52 cards so it’s replaced so it should be 3/51

step-by-step explanation: