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(D) The slope of the normal to the curve $y = f(x)$ at a point is given by $m_n = 	an(	heta)$,where $	heta$ is the angle the normal makes with the

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(D) The slope of the normal to the curve $y = f(x)$ at a point is given by $m_n = 	an(	heta)$,where $	heta$ is the angle the normal makes with the positive $x$-axis.Given $	heta = \frac{3\pi}{4}$,the slope of the normal is $m_n = 	an\left(\frac{3\pi}{4}ight) = -1$.We know that the slope of the normal is related to the slope of the tangent $(f'(x))$ by the formula $m_n = -\frac{1}{f'(x)}$.At the point $(3, 4)$,the slope of the tangent is $f'(3)$.Therefore,$-1 = -\frac{1}{f'(3)}$.Solving for $f'(3)$,we get $f'(3) = 1$.

If the normal to the curve $y = f(x)$ at the point $(3, 4)$ makes an angle $\frac{3\pi}{4}$ with the positive $x$-axis,then $f'(3)$ is equal to

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