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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 $$[0,1]$$

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