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) =

1

We have,

$y=f(x)$

$\Rightarrow \frac{d y}{d x}=f'(x) \Rightarrow\left(\frac{d y}{d x}\right)_{(3,4)}=f'(3) \Rightarrow-\frac{1}{\left(\frac{d y}{d x}\right)_{(3,4)}}=-\frac{1}{f'(3)}$

It is given that the slope of the normal $=\tan \frac{3 \pi}{4}$

∴  $\tan \frac{3 \pi}{4}=-\frac{1}{f'(3)} \Rightarrow f'(3)=1$