Find the minimum value of f(x) `=8^(x) +8^(-x) -4(4^(x)+4^(-x)) , AA x in R`

Answer

To find the minimum value of the function \( f(x) = 8^x + 8^{-x} - 4(4^x + 4^{-x}) \), we will follow these steps:

### Step 1: Rewrite the Function
We can express \( 8^x \) and \( 4^x \) in terms of \( 2^x \):
\[ 8^x = (2^3)^x = (2^x)^3 \]
\[ 4^x = (2^2)^x = (2^x)^2 \]
Let \( y = 2^x \). Then, we can rewrite \( f(x) \) as:
\[ f(x) = y^3 + \frac{1}{y^3} - 4\left(y^2 + \frac{1}{y^2}\right) \]

### Step 2: Simplify the Function
Now, we simplify \( f(x) \):
\[ f(y) = y^3 + \frac{1}{y^3} - 4\left(y^2 + \frac{1}{y^2}\right) \]
Using the identities:
\[ y^3 + \frac{1}{y^3} = \left(y + \frac{1}{y}\right)^3 - 3\left(y + \frac{1}{y}\right) \]
\[ y^2 + \frac{1}{y^2} = \left(y + \frac{1}{y}\right)^2 - 2 \]
Let \( z = y + \frac{1}{y} \). Then:
\[ f(y) = z^3 - 3z - 4(z^2 - 2) \]
\[ = z^3 - 4z^2 + 5 - 3z \]
\[ = z^3 - 4z^2 - 3z + 5 \]

### Step 3: Differentiate the Function
Now, we differentiate \( f(z) \):
\[ f'(z) = 3z^2 - 8z - 3 \]

### Step 4: Find Critical Points
To find the critical points, we set \( f'(z) = 0 \):
\[ 3z^2 - 8z - 3 = 0 \]
Using the quadratic formula:
\[ z = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{8 \pm \sqrt{(-8)^2 - 4 \cdot 3 \cdot (-3)}}{2 \cdot 3} \]
\[ = \frac{8 \pm \sqrt{64 + 36}}{6} = \frac{8 \pm \sqrt{100}}{6} = \frac{8 \pm 10}{6} \]
Calculating the two possible values:
\[ z_1 = \frac{18}{6} = 3, \quad z_2 = \frac{-2}{6} = -\frac{1}{3} \]

### Step 5: Evaluate the Second Derivative
Next, we check the second derivative to determine the nature of the critical points:
\[ f''(z) = 6z - 8 \]
Evaluating at \( z = 3 \):
\[ f''(3) = 6(3) - 8 = 18 - 8 = 10 > 0 \]
This indicates a local minimum at \( z = 3 \).

### Step 6: Calculate the Minimum Value
Now we substitute \( z = 3 \) back into the function:
\[ f(3) = 3^3 - 4(3^2) - 3(3) + 5 \]
\[ = 27 - 36 - 9 + 5 = -13 \]

Thus, the minimum value of \( f(x) \) is:
\[ \boxed{-13} \]