Answer:
B. [tex]-\frac15[/tex]
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 = [tex]-\frac15[/tex]
using calculus (disregard if you are not in calculus)
setting f(x) = y
[tex]y=\left \{ {{5x+1, x\geq -\frac15 \atop {-5x-1, x\leq -\frac15}} \right.[/tex]
[tex]\frac{dy}{dx} = \left \{ {{5, x> -\frac15 \atop {-5, x< -\frac15}} \right.[/tex]
finding critical points, where [tex]\frac{dy}{dx}[/tex] is 0 or undefined
[tex]\frac{dy}{dx}[/tex] is undefined at x = [tex]-\frac15[/tex]
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, [tex]\frac{dy}{dx}[/tex] = 5
at x = -1, [tex]\frac{dy}{dx}[/tex] = -5
since [tex]\frac{dy}{dx}[/tex] goes from negative to positive at x = [tex]-\frac15[/tex],
x = [tex]-\frac15[/tex] is a relative minimum
since x = [tex]-\frac15[/tex] is our only critical point, it is also our absolute minimum (relative minimums do not always correspond to absolute minimums)
