If the normal to the curve y=f(x) at the poin | vedclass

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(A) The slope of the normal to the curve $y=f(x)$ at a point $(x_0, y_0)$ is given by $m_n = 	an(	heta)$,where $	heta$ is the angle the normal make

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(A) The slope of the normal to the curve $y=f(x)$ at a point $(x_0, y_0)$ is given by $m_n = 	an(	heta)$,where $	heta$ is the angle the normal makes with the positive $x$-axis.Given $	heta = 3\pi/4$,the slope of the normal is $m_n = 	an(3\pi/4) = -1$.We know that the slope of the normal is also related to the derivative of the function by the formula $m_n = -1/f'(x_0)$.Substituting the given point $(3,4)$,we have $m_n = -1/f'(3)$.Equating the two expressions for the slope of the normal: $-1/f'(3) = -1$.Solving for $f'(3)$,we get $f'(3) = 1$.

$X = x_i$	$-2$	$-1$	$0$	$1$	$2$
$P(X = x_i)$	$\frac{1}{6}$	$k$	$\frac{1}{4}$	$k$	$\frac{1}{6}$

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

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