What is the normal of f(x)=sqrt(x^2+16),\at x=3 ?

The normal of f(x)=sqrt(x^2+16),\at x=3 is f(x)=-5/3 x+10

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normal of f(x)=sqrt(x^2+16),\at x=3

normal of

f (x) = x^{2} + 16, at x = 3

f (x) = - \frac{5}{3} x + 10

Solution steps

Find the tangent point:

(3, 5)

Find the slope of

f (x) = x^{2} + 16 : \frac{df ( x )}{d x} = \frac{x}{x ^{2} + 16}

EN : T i tl e G e n er a l Eq u a t i o n Sl o p e A t P o in t 2 Eq : m = \frac{3}{5}

Compute the slope of the perpendicular line:

m_{p} = - \frac{5}{3}

Find the line with slope m=

- \frac{5}{3}

and passing through

(3, 5) : f (x) = - \frac{5}{3} x + 10

f (x) = - \frac{5}{3} x + 10

What is the normal of f(x)=sqrt(x^2+16),\at x=3 ?
The normal of f(x)=sqrt(x^2+16),\at x=3 is f(x)=-5/3 x+10