The normal to the curve y=f(x) at the point ( | vedclass

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(A) The slope of the tangent to the curve $y=f(x)$ at any point is given by $\frac{dy}{dx} = f^{\prime}(x)$.Let $m_t$ be the slope of the tangent and

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(A) The slope of the tangent to the curve $y=f(x)$ at any point is given by $\frac{dy}{dx} = f^{\prime}(x)$.Let $m_t$ be the slope of the tangent and $m_n$ be the slope of the normal at the point $(3,4)$.The normal makes an angle $	heta = \frac{3\pi}{4}$ with the positive $X$-axis.Therefore,the slope of the normal is $m_n = 	an\left(\frac{3\pi}{4}ight) = -1$.We know that the relationship between the slope of the tangent and the slope of the normal is $m_n = -\frac{1}{m_t}$.Substituting the values,we get $-1 = -\frac{1}{f^{\prime}(3)}$.This implies $f^{\prime}(3) = 1$.

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 $f^{\prime}(3)$ is equal to:

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