If f: R rightarrow R is defined by f(x) = x^2 | vedclass

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(C) To determine if the function $f(x) = x^2 + 3x + 4$ is one-one or onto,we analyze its properties:$1$. One-one check: $f(x_1) = f(x_2) \implies x_1^

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many-one but not onto

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(C) To determine if the function $f(x) = x^2 + 3x + 4$ is one-one or onto,we analyze its properties:$1$. One-one check: $f(x_1) = f(x_2) \implies x_1^2 + 3x_1 + 4 = x_2^2 + 3x_2 + 4$. This simplifies to $(x_1 - x_2)(x_1 + x_2 + 3) = 0$. Since $x_1$ can be $- (x_2 + 3)$,the function is many-one.$2$. Onto check: The range of the quadratic function $f(x) = x^2 + 3x + 4$ is $[-\frac{D}{4a}, \infty)$. Here $D = b^2 - 4ac = 3^2 - 4(1)(4) = 9 - 16 = -7$. Thus,the range is $[-\frac{-7}{4}, \infty) = [1.75, \infty)$. Since the range is not equal to the codomain $R$,the function is not onto.Therefore,the function is many-one but not onto.

If $f: R \rightarrow R$ is defined by $f(x) = x^2 + 3x + 4$,then the function $f$ is . . . . . . .

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