Answer:
15.8
Step-by-step explanation:
[tex]x \in \mathbb{R}[/tex]
[tex]|5x^2 - 16| + |10x - 2|[/tex]
Once we are working with absolute values, the expression will always be positive, therefore, to get the lowest value for the expression, the lowest value for x should be 0.
[tex]|5(0)^2 - 16| + |10(0) - 2|[/tex]
[tex]|- 16| + |- 2|[/tex]
[tex]16 + 2[/tex]
[tex]18[/tex]
But this is not the right approach and this is not the lowest value. For this question, you may think that
[tex]|5x^2 - 16| + |10x - 2|>0[/tex]
For
[tex]|5x^2 - 16| + |10x - 2|=0, \nexists x \in \math{R}[/tex]
Therefore,
[tex]|5x^2 - 16| + |10x - 2|>0[/tex]
Solving that
[tex]$|5x^2-16| \implies \left|5\left(-\frac{4}{\sqrt{5}}\right)^2-16\right| = 0$[/tex]
[tex]$|10x - 2| \implies \left|10\left(\frac{1}{5} \right) - 2\right| =0$[/tex]
Once it is true for all values of x in the Real set, it means the intervals,
[tex]$x\le -\frac{4}{\sqrt{5}}$[/tex]
[tex]$-\frac{4}{\sqrt{5}}<x<\frac{1}{5}$[/tex]
[tex]$\frac{1}{5}\le x<\frac{4}{\sqrt{5}}$[/tex]
[tex]$x\ge \frac{4}{\sqrt{5}}$[/tex]
Are true and equal to [tex]x\in(-\infty, \infty)[/tex]
The lowest value for [tex]x[/tex] will be
[tex]$\frac{4}{\sqrt{5} } \text{ or } \frac{1}{5} $[/tex]
If you replace one of these values for [tex]x[/tex], you will find that [tex]$x=\frac{1}{5} $[/tex] is the value that will give the lowest value for the expression.
[tex]$\left|5\left(\frac{1}{5}\right)^2-16\right| + \left|10\left(\frac{1}{5} \right) - 2\right| = 15.8$[/tex]
