If f(x) = x^2 - 3x,then the points at which f | vedclass

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(D) Given $f(x) = x^2 - 3x$.First,find the derivative $f'(x)$:$f'(x) = \frac{d}{dx}(x^2 - 3x) = 2x - 3$.We are given the condition $f(x) = f'(x)$:$x^2

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None of these

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(D) Given $f(x) = x^2 - 3x$.First,find the derivative $f'(x)$:$f'(x) = \frac{d}{dx}(x^2 - 3x) = 2x - 3$.We are given the condition $f(x) = f'(x)$:$x^2 - 3x = 2x - 3$.Rearranging the terms to form a quadratic equation:$x^2 - 5x + 3 = 0$.Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$,where $a=1, b=-5, c=3$:$x = \frac{5 \pm \sqrt{(-5)^2 - 4(1)(3)}}{2(1)} = \frac{5 \pm \sqrt{25 - 12}}{2} = \frac{5 \pm \sqrt{13}}{2}$.Thus,the points are $\frac{5 + \sqrt{13}}{2}$ and $\frac{5 - \sqrt{13}}{2}$,which are not listed in the options.Therefore,the correct option is $D$.

If $f(x) = x^2 - 3x$,then the points at which $f(x) = f'(x)$ are

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