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(D) The slope of the tangent to the curve $y = f(x)$ at $x = 3$ is given by $f'(3)$.The slope of the normal to the curve at the point $(3, 4)$ is give

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(D) The slope of the tangent to the curve $y = f(x)$ at $x = 3$ is given by $f'(3)$.The slope of the normal to the curve at the point $(3, 4)$ is given by $m_n = -\frac{1}{f'(3)}$.Given that the normal makes an angle of $	heta = 3\pi /4$ with the $x$-axis,the slope of the normal is $m_n = 	an(3\pi /4)$.We know that $	an(3\pi /4) = -1$.Therefore,$-\frac{1}{f'(3)} = -1$.Multiplying both sides by $-1$,we get $\frac{1}{f'(3)} = 1$.Thus,$f'(3) = 1$.

If the normal to the curve $y = f(x)$ at the point $(3, 4)$ makes a positive angle of $3\pi /4$ with the $x$-axis,then $f'(3) = .....$

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