What is the equation of the line that is normal to $f (x) = x^{3} - ln (x^{2} + 1)$ $f(x)= x^3 -ln(x^2+1)$ at $x = 2$ $x= 2$ ?

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Answer 1 · 2017-03-05T21:28:46

$y = - \frac{5}{56} (x - 2) + 8 - ln 5$ $y=-5/56(x-2)+8-ln5$

$f (x) = x^{3} - ln (x^{2} + 1)$ $f(x)=x^3-ln(x^2+1)$

$f (2) = (2^{3}) - ln (2^{2} + 1) = 8 - ln 5$ $f(2)=(2^3)-ln(2^2+1)=8-ln5$

$f'(x)=3x^2-frac{2x}{x^2+1}$

$f'(2)=3(2)^2-frac{2(2)}{(2)^2+1}=12-(4/5)=56/5$

Slope of the tangent line at $(2, 8-ln5)$ is $56/5$ , and the slope of the normal line is the negative reciprocal, which is $-5/56$ .

Equation of normal line at $(2, 8-ln5)$
$y=-5/56(x-2)+8-ln5$

What is the equation of the line that is normal to f(x)= x^3 -ln(x^2+1) f(x)=x3−ln(x2+1)f(x)= x^3 -ln(x^2+1) at x= 2 x=2 x= 2 ?