The maximum value of the function f(x)=1/x^2-2|x|

Question

The maximum value of the function f(x)=1/x^2-2|x|+2 is

Deepak Kumar Shringi · Accepted Answer

To find the maximum value of the function $f (x) = \frac{1}{x^{2}} - 2 | x | + 2$ , we can break it down step by step. This function involves both a rational term and an absolute value term, which means we need to consider both cases for $x$ : positive and negative. Let's analyze the function in detail.

Step 1: Split the Function Based on Absolute Value

The absolute value function $| x |$ can be expressed differently depending on the sign of $x$ . Hence, we can rewrite $f (x)$ for two cases:

For $x \geq 0$ : $f (x) = \frac{1}{x^{2}} - 2 x + 2$
For $x < 0$ : $f (x) = \frac{1}{x^{2}} + 2 x + 2$

Step 2: Analyze Each Case

Now, let's analyze each case separately to find the maximum value.

Case 1: $x \geq 0$

We need to find the critical points by taking the derivative of $f (x)$ and setting it to zero:

First, calculate the derivative:

Let $f (x) = \frac{1}{x^{2}} - 2 x + 2$ .

Then, the derivative $f^{'} (x) = - \frac{2}{x^{3}} - 2$ .

Setting the derivative to zero:

0 = -\frac{2}{x^3} - 2

$\frac{2}{x^{3}} = - 2$ (This equation has no solutions for $x \geq 0$ since the left side is always positive.)

Next, we evaluate the function as $x$ approaches 0 and as $x$ approaches infinity:

As $x \to 0^{+}$ , $f (x) \to + \infty$ .
As $x \to \infty$ , $f (x) \to 2$ .

Case 2: $x < 0$

For this case, we have $f (x) = \frac{1}{x^{2}} + 2 x + 2$ .

Calculating the derivative:

Let $g (x) = \frac{1}{x^{2}} + 2 x + 2$ .

Then, the derivative $g^{'} (x) = - \frac{2}{x^{3}} + 2$ .

Setting the derivative to zero:

0 = -\frac{2}{x^3} + 2

$\frac{2}{x^{3}} = 2$ implies $x^{3} = 1$ , so $x = - 1$ (since we are considering $x < 0$ ).

Now, we evaluate $g (- 1)$ :

g(-1) = \frac{1}{(-1)^2} + 2(-1) + 2 = 1 - 2 + 2 = 1.

Step 3: Comparing Values

Now, let's summarize the findings:

For $x \geq 0$ , $f (x)$ approaches infinity as $x$ approaches 0 and approaches 2 as $x$ approaches infinity.
For $x < 0$ , the maximum value found is $1$ at $x = - 1$ .