Yet to be titled...

Calculus Level 3

f {f} ( x {x} ) is a quadratic function with one real root such that: f ( x ) = 2 x 2 + b x − c f(x) = 2x^2 +bx -c f o r b , c ∈ ℜ for \ b,c\in \Re What is the minimum value of 4 b + c 2 ? 4b + c^2 \ ?


The answer is -12.

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Curtis Clement by Curtis Clement , 23, United Kingdom

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1 solution

Curtis Clement
Feb 8, 2015

Using the discriminant: b 2 − 4 a c = b 2 + 8 c = 0 ⇒ c = − b 2 8 b^2 - 4ac = b^2 + 8c = 0 \Rightarrow\ c = -\frac{b^2}{8} 4 b + c 2 = 4 b + b 4 64 = 256 b + b 4 64 = y 4b + c^2 = 4b +\frac{b^4}{64}\ = \frac{256b + b^4}{64}\ = y Now we must find the minimum value of y {y} by using differentiation as follows: d y d b = 4 + b 3 16 = 0 ⇒ b = − 4 ⇒ d 2 y d b 2 = 3 b 2 16 = 3 > 0 \frac{dy}{db}\ = 4 + \frac{b^3}{16} = 0\Rightarrow\ b = -4 \Rightarrow \frac{d^2 y}{db^2}\ = \frac{3b^2}{16}\ = 3 > 0 So the minimum value occurs at: b = − 4 ∴ 4 b + c 2 ≥ − 12 b = -4 \therefore\ 4b + c^2\geq\ -12

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I don't understand why 4b - c^{2} = 4b + c^{2}

Ahmed Kamel - 6 years, 4 months ago

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Sorry it's a typo - I've fixed it now

Curtis Clement - 6 years, 4 months ago

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