Question

PLEASE HELP????
What is the minimum of 2 x^2+2^{5} x^{-2}?

Answer 1

Let's start by finding the critical points of the function.
\[
f(x) = 2x^2 + 2^5x^{-2} \\
f'(x) = 4x -2^6x^{-3} \\
f'(x) = 0 \quad \to \quad 4x -2^6x^{-3} = 0 \\
4x^4 - 2^6 = 0 \\
4x ^4 = 2^6 \\
x^4 = \frac{2^6}{4} = \frac{2^6}{2^2} = 2^4 \\
x = \pm\; 2 \\
\]
Now for each let's check the second derivative to see if they are minimum or maximum.
\[
f''(x) = \bigg[f'(x) \bigg]' = \bigg[4x -2^6x^{-3}\bigg]' = 4 + 3 \cdot 2^{6}x^{-4}\\
\]
\(f''(x) > 0 \) Means the slope of \(f(x)\) is increasing at \(x\), and on a critical point it means it's a minimum point.
\[
f''(2) = 4 + 3\cdot 2^6 \cdot 2^{-4} = 4 + 3 \cdot2^2 = 16\\
f''(-2) = 4 + 3\cdot 2^6 \cdot (-2)^{-4} = 4 + 3 \cdot2^2 = 16
\]
Means... both \(x=2\) and \(x=-2\) are minimum points of \(f(x)\) at value of 16.
For reference:
http://www.wolframalpha.com/input/?i=y+%3D+2x%C2%B2+%2B+32x%E2%81%BB%C2%B2
Agrees about that as well