The coordinates of the point(s) on the graph | vedclass

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(D) Given the function $f(x) = \frac{x^3}{3} - \frac{5x^2}{2} + 7x - 4$. The slope of the tangent at any point $(x, y)$ is given by $\frac{dy}{dx} = x

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Both $(A)$ and $(B)$

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(D) Given the function $f(x) = \frac{x^3}{3} - \frac{5x^2}{2} + 7x - 4$. The slope of the tangent at any point $(x, y)$ is given by $\frac{dy}{dx} = x^2 - 5x + 7$. Since the intercepts are equal in magnitude but opposite in sign,the slope of the tangent must be $-1$ or $1$. Case $1$: Slope $m = -1$. $x^2 - 5x + 7 = -1 \implies x^2 - 5x + 8 = 0$. The discriminant $D = 25 - 32 = -7 Case $2$: Slope $m = 1$. $x^2 - 5x + 7 = 1 \implies x^2 - 5x + 6 = 0 \implies (x-2)(x-3) = 0$. Thus,$x = 2$ or $x = 3$. For $x = 2$,$f(2) = \frac{8}{3} - \frac{5(4)}{2} + 7(2) - 4 = \frac{8}{3} - 10 + 14 - 4 = \frac{8}{3}$. Point is $(2, 8/3)$. For $x = 3$,$f(3) = \frac{27}{3} - \frac{5(9)}{2} + 7(3) - 4 = 9 - 22.5 + 21 - 4 = 3.5 = 7/2$. Point is $(3, 7/2)$. Both points satisfy the condition. Therefore,the correct option is $(D)$.

The coordinates of the point$(s)$ on the graph of the function $f(x) = \frac{x^3}{3} - \frac{5x^2}{2} + 7x - 4$ where the tangent drawn cuts off intercepts from the coordinate axes which are equal in magnitude but opposite in sign,are:

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