A function f: R rightarrow R defined by f(x) | vedclass

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(B) To determine the nature of the function $f(x)$,we analyze its behavior in two intervals.For $x \leq \frac{4}{3}$,$f(x) = 2x+3$. This is a strictly

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not onto

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(B) To determine the nature of the function $f(x)$,we analyze its behavior in two intervals.For $x \leq \frac{4}{3}$,$f(x) = 2x+3$. This is a strictly increasing linear function. The range for this part is $(-\infty, 2(\frac{4}{3})+3] = (-\infty, \frac{17}{3}]$.For $x > \frac{4}{3}$,$f(x) = -3x^2+8x$. This is a downward-opening parabola with vertex at $x = -\frac{b}{2a} = -\frac{8}{2(-3)} = \frac{4}{3}$. Since the interval is $x > \frac{4}{3}$,the function is strictly decreasing in this domain. The value at $x = \frac{4}{3}$ is $-3(\frac{16}{9}) + 8(\frac{4}{3}) = -\frac{16}{3} + \frac{32}{3} = \frac{16}{3}$. As $x \rightarrow \infty$,$f(x) \rightarrow -\infty$. Thus,the range for this part is $(-\infty, \frac{16}{3})$.Combining these,the function is not one-one because the value $\frac{16}{3}$ is attained at $x = \frac{4}{3}$ and also at some $x > \frac{4}{3}$ (since the parabola decreases from $\frac{16}{3}$). Since the range is $(-\infty, \frac{17}{3}]$,which is not equal to the codomain $R$,the function is not onto. Therefore,it is not bijective. The correct description is that it is not onto.

$A$ function $f: R \rightarrow R$ defined by $f(x) = \begin{cases} 2x+3, & x \leq \frac{4}{3} \\ -3x^2+8x, & x > \frac{4}{3} \end{cases}$ is

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