The function's minimum value is -2 at x = 1
Step-by-step explanation:
Given function is:
[tex]y = 2x^2+4x[/tex]
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
[tex]\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[/tex]
We have to find the roots of first derivative
So,
[tex]4x-4 = 0\\4x = 4\\x = \frac{4}{4} \\x = 1[/tex]
Taking second derivative
[tex]\frac{d}{dx} (y') = \frac{d}{dx} (4x-4)\\y'' = 4[/tex]
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
[tex]= 2(1)^2 - 4(1)\\= 2(1) - 4\\= 2-4\\= -2[/tex]
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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