Evaluate the functions `f(x)=x^(2)+1/x^(2) and phi(x)=x^4+1/x^4`
for the points at which `1/x +x=5`

Answer

To solve the problem, we need to evaluate the functions \( f(x) = x^2 + \frac{1}{x^2} \) and \( \phi(x) = x^4 + \frac{1}{x^4} \) at the points where \( \frac{1}{x} + x = 5 \).

### Step 1: Solve for \( x \) from the equation \( \frac{1}{x} + x = 5 \)

First, we can rewrite the equation:

\[ x + \frac{1}{x} = 5 \]

Multiplying both sides by \( x \) (assuming \( x \neq 0 \)) gives:

\[ x^2 + 1 = 5x \]

Rearranging this gives:

\[ x^2 - 5x + 1 = 0 \]

### Step 2: Use the quadratic formula to find \( x \)

The quadratic formula is given by:

\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

For our equation \( x^2 - 5x + 1 = 0 \), we have \( a = 1, b = -5, c = 1 \). Plugging in these values:

\[ x = \frac{5 \pm \sqrt{(-5)^2 - 4 \cdot 1 \cdot 1}}{2 \cdot 1} \]

Calculating the discriminant:

\[ x = \frac{5 \pm \sqrt{25 - 4}}{2} \]
\[ x = \frac{5 \pm \sqrt{21}}{2} \]

### Step 3: Calculate \( f(x) \)

Now we need to evaluate \( f(x) = x^2 + \frac{1}{x^2} \). We can use the identity:

\[ x^2 + \frac{1}{x^2} = \left( x + \frac{1}{x} \right)^2 - 2 \]

Since \( x + \frac{1}{x} = 5 \):

\[ f(x) = 5^2 - 2 = 25 - 2 = 23 \]

### Step 4: Calculate \( \phi(x) \)

Next, we need to evaluate \( \phi(x) = x^4 + \frac{1}{x^4} \). We can use the identity:

\[ x^4 + \frac{1}{x^4} = \left( x^2 + \frac{1}{x^2} \right)^2 - 2 \]

From the previous step, we found \( x^2 + \frac{1}{x^2} = 23 \):

\[ \phi(x) = 23^2 - 2 = 529 - 2 = 527 \]

### Final Answer

Thus, the values are:

\[ f(x) = 23 \quad \text{and} \quad \phi(x) = 527 \]