### Instruction: What is the range of the function $$F(x) = |x+1|+|x-5|~?$$Express your answer in interval notation.### Response: I know that the absolute value function always returns a non-negative value, regardless of the sign of the input.
So, I can say that $$F(x) \geq 0$$ for any value of x.
But is there a lower bound for F(x)?
I can try to find the minimum value of F(x) by considering different cases for the signs of the two terms inside the absolute value bars.
If x is less than -1, then both x+1 and x-5 are negative, so F(x) = -(x+1)-(x-5) = -2x+4.
If x is between -1 and 5, then x+1 is positive and x-5 is negative, so F(x) = (x+1)-(x-5) = 6.
If x is greater than 5, then both x+1 and x-5 are positive, so F(x) = (x+1)+(x-5) = 2x-4.
I can see that F(x) is a linear function in each of these cases, and it changes slope at the points x = -1 and x = 5.
I can also see that F(x) is decreasing when x < -1, constant when -1 < x < 5, and increasing when x > 5.
This means that the minimum value of F(x) occurs when x = -1 or x = 5, and in both cases F(x) = 6.
So, the lowest possible value of F(x) is 6, and there is no upper bound for F(x).
Therefore, the range of F(x) is the interval [6, infinity).