F-R - R is defined as fx - x4 - 6x2 - 12- The range of fx is

The correct option is C [3, ∞) f(x)=x^4 6x^2+12 f(x) = (x^2)^2 2 × x^2 × 3 + 3^2 +3 f(x) = (x^2 3)^2+3 (x^2 3)^2 ≥ 0 (x^2 3)^2+3 ≥ 3 ⇒ f(x) ≥ 3 The range o ...

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Range

f:R → R is de...

Question

$f : R \to R$ is defined as $f (x) = x^{4} - 6 x^{2} + 12$ . The range of f(x) is

$(- \infty, \infty)$

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$[- 3, \infty)$

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$[3, \infty)$

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$[12, \infty)$

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Solution

The correct option is C
$[3, \infty)$

$f (x) = x^{4} - 6 x^{2} + 12 f (x) = (x^{2})^{2} - 2 \times x^{2} \times 3 + 3^{2} + 3 f (x) = (x^{2} - 3)^{2} + 3 (x^{2} - 3)^{2} \geq 0 (x^{2} - 3)^{2} + 3 \geq 3 \Rightarrow f (x) \geq 3$
The range of f(x) is $[3, \infty)$

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f:R→R is defined as f(x)=x4−6x2+12. The range of f(x) is

The correct option is C [3,∞) f(x)=x4−6x2+12f(x)=(x2)2−2×x2×3+32+3f(x)=(x2−3)2+3(x2−3)2≥0(x2−3)2+3≥3⇒f(x)≥3 The range of f(x) is [3,∞)

$f : R \to R$ is defined as $f (x) = x^{4} - 6 x^{2} + 12$ . The range of f(x) is

The correct option is C
$[3, \infty)$

$f (x) = x^{4} - 6 x^{2} + 12 f (x) = (x^{2})^{2} - 2 \times x^{2} \times 3 + 3^{2} + 3 f (x) = (x^{2} - 3)^{2} + 3 (x^{2} - 3)^{2} \geq 0 (x^{2} - 3)^{2} + 3 \geq 3 \Rightarrow f (x) \geq 3$
The range of f(x) is $[3, \infty)$