If the minimum value

$\quad f(x) = |x+1|+|ax+1| \\ \quad \text{Critical points : } -\dfrac{1}{a} \quad , \quad -1 \\ \quad f(-\frac{1}{a}) = \left| 1-\dfrac{1}{a} \right| = 1.5 \implies a=\dfrac{2}{5} \quad , \quad -2 \\ \quad f(-1) = | 1-a | = 1.5 \implies a=\dfrac{5}{2} \quad , \quad -\dfrac12 \\ \quad \text{These are possible values of a , we have to check if they are correct or not }\\ \quad \text{For } a=\dfrac{5}{2} \\ \quad f(x)= |x+1|+|\dfrac{5}{2}x+1| \\ \quad f(-1)= 1.5 \quad , \quad f\left(-\frac{2}{5}\right)= \dfrac{3}{5} \quad \color{royalblue}{\text{Minimum is not 1.5}} \\ . \\ \quad \text{For } a=\dfrac{2}{5} \\ \quad f(x)= |x+1|+|\dfrac{2}{5}x+1| \\ \quad f(-1)= \dfrac{3}{5} \quad , \quad f\left(-\frac{5}{2}\right)= -1 \quad \color{royalblue}{\text{Minimum is not 1.5}} \\ . \\ \quad \text{For } a= -2 \\ \quad f(x)= |x+1|+|1-2x| \\ \quad f(-1)= 3 \quad , \quad f\left(\dfrac{1}{2}\right)= \dfrac{3}{2} \quad \quad \color{royalblue}{\text{Minimum is 1.5}} \\ .\\ \quad \text{For } a= -\dfrac{1}{2} \\ \quad f(x)= |x+1|+|1-\dfrac12x| \\ \quad f(-1)= \dfrac32 \quad , \quad f\left(2\right)= 3 \quad \quad \color{royalblue}{\text{Minimum is 1.5}} \\ \quad \text{ Hence } a= -2 , -0.5 \\ \quad \text{Our answer: } \boxed{-2.5}$

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$\text{Nice Question !}$

First we observe that the graph of $|x+1|$ and we see that, at x=-1, it reaches it's minimum, 0.

So, to solve this, we consider a solution for which, at $x=-1, |x+1| = 1.5$

There are 2 cases:

$Case 1. -a+1=1.5$

$a=-.5$

$Case 2. -a+1=-1.5$

$a=2.5$

Perhaps we check to see if this works, and find that the first case works and the second does not. The reason for this is that the a does not necessarily minimize at -1. It only does this when the second expression's x can't overpower it, which it will not only if a is between 1 and -1.

So we have 1 solution, $a=-5$ .

But what do we do about solutions in which the ax does overpower it the x and move the minimum?

Because of the linear nature of these expressions, the minimum must be at a point of change. We got the first point of change above, by looking at the graph of $x+1$ . Using these observations we can figure out that $ax+1$ , minimizes at 0, at that the point where it reaches it's minimum is therefore $ax+1=0$

$ax=-1$

$x=-1/a$

Imagining the graph of the equation that is given to us with this information assuming that $ax>|1|$ , there will be two lines coming down that will make sharp turns at $x=-1 and x=-1/a.$ In between $x=-1 and x=-1/a$ it will decrease at a rate of $a-1$ (since the two lines will be 'fighting' each other, the ax going down and the x going up). So the total decrease from the point x=-1 to the point x=-1/a will be equal to the distance between the points $(1-1/a)$ times the steepness of the slope between the points $(a-1)$ . At the point x=-1 we can figure out the height of the graph in terms of a.

$f(x) = |x+1| + |ax+1|$

$f(x) = |-1+1| + |a*-1+1|$

$f(x) = |-a+1|$

Now, we know that the minimum is equal to this height minus the change between the point x=-1 and x=-1/a, which I figured out in the last paragraph. We can now write our final equation to find the minimum based on a (when a is not between -1 and 1)! EXCITING!

$|-a+1| - (a-1)(1-1/a)$

Let's set that equal to 1.5, our goal, and solve.

$|-a+1| - (a-1)(1-1/a)=1.5$

$|-a+1| - (a-2+1/a)=1.5$

$|-a+1| -a + 2 - 1/a=1.5$

$|-a+1| -a + .5 - 1/a=0$

Now we'll split it into the 2 cases...

$Case 1. -a+1=-a + .5 - 1/a$

$.5= - 1/a$

$a = -2$

$Case 2. a-1=-a + .5 - 1/a$

$0=-2a+1.5-1/a$

$0=-2a^2+1.5a-1$

The quadratic formula gives us imaginary solutions for Case 2, but for Case 1 we get a nice, non-imaginary answer of -2! Add that to our previous answer of -.5 and, at last, the solution-> -2.5!

I tried to make this solution easy to follow, did I do a good job? Pretty much the first time I used latex, too. It wasn't as hard as I thought. The solution might not be "BRILLIANT!" but, hey, I solved it, right?