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• Notes

Find an equation of the line normal to the graph of $y=x^{2}+\frac{x^{2}-7}{2 x+1}$ at $x=1$

$y=x^{2}+\frac{x^{2}-7}{2 x+1}, \text { at } x=1$

$y=(1)^{2}+\frac{(1)^{2}-7}{2(1)+1}=1+\frac{-6}{3}=1-2=-1$

$y^{\prime}=2 x+\frac{2 x(2 x+1)-(2 x)\left(x^{2}-7\right)}{(2 x+1)^{2}} \text { at } x=1$

$y^{\prime}=2(1)+\frac{2(1)(2(1)+1)-(2)(1)((1)^{2}-7)}{(2(1)+1)^{2}}$

$=2+\frac{6+12}{9}$

$y^{\prime}=\frac{18+18}{9}=4$

$m=\frac{-1}{4} \quad p(1,-1)$

$y-y_{0}=m(x-x_{0})$

$y+1=\frac{-1}{4}(x-1)$

Find Equation for tangent and normal line of $y=x^{4}+2 e^{x}$ at $(0,2)$

Equation of tangents $\rightarrow y-y_{0}=m\left(x-x_{0}\right)$

$m=f^{\prime}(0) \rightarrow f^{\prime}(x)=4 x^{3}+2 e^{x} \rightarrow f^{\prime}(0)=4(0)^{3}+2 e^{0}=2$

Equation of tangent: $(y-2)=2(x-0) \rightarrow y-2=2 x$

$y=2 x+2$

Equation of normal line $\rightarrow \quad y-y_{0}=\frac{-1}{f^{\prime}(x)}\left(x-x_{0}\right)$

$m=\frac{-1}{f^{\prime}(0)}$

$y-2=\frac{-1}{2}(x-0) \rightarrow y=\frac{-1}{2} x+2$

Show that $f(x)=\frac{x^{3}-x+1}{x^{2}+1}$ has a horizontal tangent in $[0,1]$

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

$f^{\prime}(x)=\frac{\left(3 x^{2}-1\right)\left(x^{2}+1\right)-(2 x)\left(x^{3}-x+1\right)}{\left(x^{2}+1\right)^{2}}$

$f^{\prime}(0)=\frac{(-1)(1)-0}{1}=-1<0$

$f^{\prime}(1)=\frac{(2)(2)-(1)(2)}{4}=\frac{4-2}{4}=\frac{1}{2}>0$

$f^{\prime}(x)$ cont on $[0,1]$

$f^{\prime}(0)<0<f^{\prime}(1)$

$c \in(0,1)$ such that $f^{\prime}(c)=0$ IvT

$f$ has a h.t in $[0,1]$