What is the normal of f(x)=log_{2}(2x^2+14),\at x=5 ?

The normal of f(x)=log_{2}(2x^2+14),\at x=5 is f(x)=-(16ln(2))/5 x+6+16ln(2)

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normal of f(x)=log_{2}(2x^2+14),\at x=5

normal of

f (x) = lo g_{2} (2 x^{2} + 14), at x = 5

f (x) = - \frac{16 ln ( 2 )}{5} x + 6 + 16 ln (2)

Solution steps

Find the tangent point:

(5, 6)

Find the slope of

f (x) = lo g_{2} (2 x^{2} + 14) : \frac{df ( x )}{d x} = \frac{2 x}{ln ( 2 ) ( x ^{2} + 7 )}

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{5}{16 ln ( 2 )}

Compute the slope of the perpendicular line:

m_{p} = - \frac{16 ln ( 2 )}{5}

Find the line with slope m=

- \frac{16 ln ( 2 )}{5}

and passing through

(5, 6) : f (x) = - \frac{16 ln ( 2 )}{5} x + 6 + 16 ln (2)

f (x) = - \frac{16 ln ( 2 )}{5} x + 6 + 16 ln (2)

What is the normal of f(x)=log_{2}(2x^2+14),\at x=5 ?
The normal of f(x)=log_{2}(2x^2+14),\at x=5 is f(x)=-(16ln(2))/5 x+6+16ln(2)