Set f(x)=Ax2+Bx+C and g(x)=ax2+bx+c with A×a≠0, A, aB, b, C, c∫R satisfies:|f(x)|≥|g(x)|∀x∈RProve that |B2−4AC|≥|b2−4ac|

"Set f(x)=Ax2+Bx+C and g(x)=ax2+bx+c with A×a≠0, A, aB,..." was answered by our experts

Jasper Dougherty

2022-04-02

Set

f (x) = A x^{2} + B^{x} + C

and

g (x) = a x^{2} + b x + c

with

A \times a \neq 0, A, a B, b, C, c \int R

satisfies:

| f (x) | \geq | g (x) | \forall x \in R

Prove that

| B^{2} - 4 A C | \geq | b^{2} - 4 a c |

Cody Johns

Beginner2022-04-03Added 8 answers

Step 1
The extrema value of g(x) is

\frac{b^{2} - 4 a c}{4 a}

Wrong claim that you made: The smallest value of

} g (x) ∣

need not be

| \frac{b^{2} - 4 a c}{4 a} |

E.g. Consider

g (X) = (x - 1) (x + 1)

. Clearly the smallest value of

| g (x) |

is o.
Whereas

| \frac{b^{2} - 4 a c}{4 a} |

is the absolute value of the extrema of g(x), so this is equal to

| - 1 | = 1

Step 2
You need

δ_{g} \geq 0

in order to conclude that "The smallest value of

| g (x) |

| \frac{b^{2} - 4 a c}{4 a} |

'' for your proof to work.
Note that the case of

δ_{g} \geq 0

is pretty simple to deal with.

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