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Prove that y=x² at x=0 has both a local and global minimum using first derivative test. I can prove it with a graph looking at it visually but how would you prove it with written words.

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Lidpras

[tex] \frac{d}{dx} {x}^{2} = 2x \\ x = 0 \\ 2 \times 0 = 0[/tex]

When the derivative equals zero, that point is either the maximum or minimum.

[tex] {x}^{2} \geqslant 0[/tex]

The smallest real value of x² is 0, which is at x = 0.

Hence, x² at x = 0 has both a local and global minimum.

MathPhys

Step-by-step explanation:

y = x²

Find the first derivative:

dy/dx = 2x

Find the critical values by setting dy/dx to 0:

0 = 2x

x = 0

Evaluating the first derivative before and after x=0, we see that dy/dx changes signs from negative to positive.

x < 0, dy/dx < 0

x > 0, dy/dx > 0

That means x=0 is a local minimum.

To find the global minimum, we need evaluate the function at x=0 and at the end points.

x = -∞, y = ∞

x = 0, y = 0

x = ∞, y = ∞

So x=0 is also the global minimum.

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