Question

Identify the minimum value of the function y = 2x^2 + 4x

Answer 1

The function's minimum value is -2 at x = 1

Step-by-step explanation:

Given function is:

$y = 2x^2+4x$

In order to find the maxima or minima of a function, first of all, we have to find the derivative of the function

So,

Taking derivatives on both sides

$\frac{d}{dx}(y) = \frac{d}{dx} (2x^2-4x)\\\frac{dy}{dx} = \frac{d}{dx} (2x^2) - \frac{d}{dx} (4x)\\y' = 2 \frac{d}{dx} (x^2) - 4 \frac{d}{dx}(x)\\y' = 2 (2x) \frac{d}{dx}(x) - 4(1)\\y' = 4x (1) - 4\\y' = 4x - 4$

We have to find the roots of first derivative

So,

$4x-4 = 0\\4x = 4\\x = \frac{4}{4} \\x = 1$

Taking second derivative

$\frac{d}{dx} (y') = \frac{d}{dx} (4x-4)\\y'' = 4$

We can see that the second derivative of the function is a constant so the function has a minima at x=1

For minimum value, putting x = 1 in the function

$= 2(1)^2 - 4(1)\\= 2(1) - 4\\= 2-4\\= -2$

The function's minimum value is -2 at x=1

Keywords: Functions, minimum value

Learn more about functions at:

• brainly.com/question/9880052
• brainly.com/question/9862781

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