How many methods do you know to blow this off?

Let $f(x) = |3x+8|-|4x|$ . By the definition of absolute value , we have the following $3$ cases:

$\begin{cases} \text{For } x < - \frac{8}{3} & f(x) = -3x - 8 + 4x & \Rightarrow f(x) = x - 8 & \Rightarrow f_{max_1}(x) = - 10\frac{2}{3} \\ \text{For } - \frac{8}{3} \le x < 0 & f(x) = 3x + 8 + 4x & \Rightarrow f(x) = 7x + 8 & \Rightarrow f_{max_2}(x) = 8 \\ \text{For } x \ge 0 & f(x) = 3x + 8 - 4x & \Rightarrow f(x) = 8 - x & \Rightarrow f_{max_3}(x) = 8 \end{cases}$

Therefore, the maximum value $f_{max}(x) = f_{max_2}(x) = f_{max_3}(x) = \boxed{8}$ .

The plot of $f(x)$ is as below.

$|3x + 8| - |4x |$ Case I : $x \geq 0$
$= 3x + 8 - 4x$ $= 8 - x$ Therefore, maximum value in this case is 8 when $x = 0$ .

Case II : $\dfrac{-8}{3} < x < 0$ $= 3x + 8 + 4x$ $= 7x + 8$ Maximum value in this case will be slightly less than 8 when $x \rightarrow 0^-$ .

Case III : $x \leq \dfrac{-8}{3}$ $= -3x - 8 + 4x$ $= x - 8$ Maximum value in this case is $\dfrac{-32}{3}$ when $x = \dfrac{-8}{3}$

Taking intersection of all the cases, Maximum value is $\boxed{8}$

By repeated uses of the triangle inequality,

$|3x+8| - |4x| \leq |3x+8 -4x| = |8-x| \leq 8 + |x|$ .

Thus $|3x+8| - |4x|$ is bounded above by $8 + |x|$ . Does it attain the minimum of this bound, which is $8$ ? Yes, when $x=0$ , we have $|3(0)+8| - |4(0)| = |8| = 8$ , and we're done.

we know that for the given expression |3x+8| -|4x|, we will get the maximum value when the second term inside the modulus is negative, as - * - is +. so we take 4x to be less than 0 which gives x as 0. putting x=0 in the original inequality we get the maximum value as 8. hence we got the answer.

For those who are out of paper (possibly calc)

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Think of the numbers $3$ & $4$ as a multiplier to the variable $x$ . For positive $x$ , $4x$ is always greater than $3x$ , which when subtracted from $3x$ subtracts a positive value from 8.

Now, when $x$ is negative, $4x$ is always subtracted from the equation; also, $3x$ is subtracted from $8$ initially, which leads to a more negative number than the former case. Therefore, for both cases ( $x < 0$ & $x > 0$ ), the equation is less than $8$ .

Now take $x=0$ , you see it outputs $8$ (max).

using the triangle inequality:

|3x+8|-|4x| =< |3x+8-4x|

|3x+8|-|4x| =< |8-x| =< |8|-|x|

for the inequality to satisfy the constraints, the domain of x is -9 < |x| < 9

lastly, following the logic that the less we subtract, the higher the value of the right most part of the inequality. we can then deduce that x=0 and |8| - |0| = 8

When modulus is applied to 3x + 8 and 4x . Whatever happens in 1st and 4th quadrant it 's mirror image will be formed in 2nd and 3rd quadrant . So it is enough if we concentrate on 1st and 4th quadrant. So removing modulus the equation becomes 3x + 8 -4x = 8-x. Implies the graph is maximum when x=0 . 8-0=8.

When X >= -8/3, The equations is 8-x. The largest value of this is 8. When X< -8/3, The equation is X-8. The largest value is negative(less than 8)