Questions: Find the x-values of all points where the function has any relative extrema. Find the value(s) of any relative extrema. f(x)=x^4-162 x^2-7 First find the derivative of f(x). f'(x)=

Solution Steps

The function is given by

\[ f(x) = x^4 - 162x^2 - 7. \]

To find the critical points, we first compute the derivative:

\[ f'(x) = 4x^3 - 324x. \]

Next, we set the derivative equal to zero to find the critical points:

\[ 4x^3 - 324x = 0. \]

Factoring out \(4x\), we have:

\[ 4x(x^2 - 81) = 0. \]

This gives us the critical points:

\[ x = 0, \quad x^2 - 81 = 0 \implies x = \pm 9. \]

Thus, the critical points are

\[ x = -9, \quad x = 0, \quad x = 9. \]

To classify these critical points, we compute the second derivative:

\[ f''(x) = 12x^2 - 324. \]

Now we evaluate the second derivative at each critical point:

1. For \(x = -9\):

\[ f''(-9) = 12(-9)^2 - 324 = 12 \cdot 81 - 324 = 972 - 324 = 648 > 0 \quad \text{(minimum)}. \]

2. For \(x = 0\):

\[ f''(0) = 12(0)^2 - 324 = -324 < 0 \quad \text{(maximum)}. \]

3. For \(x = 9\):

\[ f''(9) = 12(9)^2 - 324 = 12 \cdot 81 - 324 = 972 - 324 = 648 > 0 \quad \text{(minimum)}. \]

Now we substitute the critical points back into the original function to find the corresponding \(y\)-values:

1. For \(x = -9\):

\[ f(-9) = (-9)^4 - 162(-9)^2 - 7 = 6561 - 162 \cdot 81 - 7 = 6561 - 13122 - 7 = -6568. \]

2. For \(x = 0\):

\[ f(0) = 0^4 - 162 \cdot 0^2 - 7 = -7. \]

3. For \(x = 9\):

\[ f(9) = (9)^4 - 162(9)^2 - 7 = 6561 - 13122 - 7 = -6568. \]

Final Answer