Consider the function f(x) = x^3 - 8x^2 + 20x | vedclass

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(B) Given $f(x) = x^3 - 8x^2 + 20x - 13$. First,find the derivative $f'(x) = 3x^2 - 16x + 20$. Setting $f'(x) = 0$,we get $3x^2 - 16x + 20 = 0$. Using

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many-one and onto

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(B) Given $f(x) = x^3 - 8x^2 + 20x - 13$. First,find the derivative $f'(x) = 3x^2 - 16x + 20$. Setting $f'(x) = 0$,we get $3x^2 - 16x + 20 = 0$. Using the quadratic formula,$x = \frac{16 \pm \sqrt{256 - 240}}{6} = \frac{16 \pm 4}{6}$. So,$x_1 = 2$ and $x_2 = \frac{10}{3}$. Since $f'(x) > 0$ for $x  10/3$,and $f'(x) Since the range of a cubic polynomial is $(-\infty, \infty)$,the function is onto. Now,calculate $f(2) = 8 - 32 + 40 - 13 = 3$. Calculate $f(10/3) = (1000/27) - 8(100/9) + 20(10/3) - 13 = \frac{1000 - 2400 + 1800 - 351}{27} = \frac{49}{27}$. Since $f(2) \cdot f(10/3) = 3 \cdot \frac{49}{27} = \frac{49}{9} > 0$,option $D$ is incorrect. Since $f(2) > 0$ and $f(10/3) > 0$,and the local minimum is positive,the graph crosses the $x$-axis only once. Thus,it has only $1$ real root. Therefore,the function is many-one and onto.

Consider the function $f(x) = x^3 - 8x^2 + 20x - 13$. The function $f: R \rightarrow R$ is:

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