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What is the local extrema/ absolute of the polynomial 10x^5 + 22x^4 + 20x^3 -0x^2 + 6?

What is the local extrema/ absolute of the polynomial 10x^5 + 22x^4 + 20x^3 -0x^2 + 6?-example-1
User MarekLi
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1 Answer

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o find the local extrema of the polynomial, we first take the derivative of the polynomial and set it equal to zero:

f(x) = 10x^5 + 22x^4 + 20x^3 - 0x^2 + 6

f'(x) = 50x^4 + 88x^3 + 60x^2

Setting f'(x) = 0, we can factor out a 2x^2 from the expression to get:

f'(x) = 2x^2(25x^2 + 44x + 30) = 0

Using the quadratic formula to solve for x, we get:

x = (-44 ± sqrt(44^2 - 4(25)(30))) / (2(25))

x = (-44 ± sqrt(784)) / 50

x = (-44 ± 28) / 50

x = -0.32 or -0.56

User Matej Ukmar
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