Question

Let x be a real number. Find the lowest possible value of |5x^2 - 16| + |10x - 2|.

Answer 1

Answer:

15.8

Step-by-step explanation:

$x \in \mathbb{R}$

$|5x^2 - 16| + |10x - 2|$

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.

$|5(0)^2 - 16| + |10(0) - 2|$

$|- 16| + |- 2|$

$16 + 2$

$18$

But this is not the right approach and this is not the lowest value. For this question, you may think that

$|5x^2 - 16| + |10x - 2|>0$

For

$|5x^2 - 16| + |10x - 2|=0, \nexists x \in \math{R}$

Therefore,

$|5x^2 - 16| + |10x - 2|>0$

Solving that

$|5x^2-16| \implies \left|5\left(-\frac{4}{\sqrt{5}}\right)^2-16\right| = 0$

$|10x - 2| \implies \left|10\left(\frac{1}{5} \right) - 2\right| =0$

Once it is true for all values of x in the Real set, it means the intervals,

$x\le -\frac{4}{\sqrt{5}}$

$-\frac{4}{\sqrt{5}}$

$\frac{1}{5}\le x$

$x\ge \frac{4}{\sqrt{5}}$

Are true and equal to $x\in(-\infty, \infty)$

The lowest value for $x$ will be

$\frac{4}{\sqrt{5} } \text{ or } \frac{1}{5}$

If you replace one of these values for $x$, you will find that $x=\frac{1}{5}$ is the value that will give the lowest value for the expression.

$\left|5\left(\frac{1}{5}\right)^2-16\right| + \left|10\left(\frac{1}{5} \right) - 2\right| = 15.8$

Answer 2

Answer:2

Step-by-step explanation: