Problems from Normal straight line to a curve at a point

Development:

a) Two possible candidates are: $f(x)=x^2+4x-3$ and $g(x)=-2x+5$.

b) First we look for the slope of $r(x)$.

Slope of $g(x):g^{\prime }(x)=-2$

$r(x)$ normal to $g(x)\to r^{\prime }(x)=-{\displaystyle \frac{1}{-2}}={\displaystyle \frac{1}{2}}$

Now we should look for the for the point where the derivative of $f(x)$ is ${\displaystyle \frac{1}{2}}$. This is $$f^{\prime }(a)=2a+4={\displaystyle \frac{1}{2}}\Rightarrow a=-{\displaystyle \frac{7}{4}}$$ $$f(-{\displaystyle \frac{7}{4}})=(-{\displaystyle \frac{7}{4}}{)}^2+4(-{\displaystyle \frac{7}{4}})-3=-{\displaystyle \frac{111}{16}}$$

The touching point will be $(a,f(a))=(-{\displaystyle \frac{7}{4}},-{\displaystyle \frac{111}{16}})$

The equation of the straight line is written $r(x)$: $$y+{\displaystyle \frac{111}{16}}={\displaystyle \frac{1}{2}}\cdot (x+{\displaystyle \frac{7}{4}})$$ $$r(x)={\displaystyle \frac{1}{2}}\cdot x-{\displaystyle \frac{97}{16}}$$

Solution:

a) $f(x)=x^2+4x-3$, $g(x)=-2x+5$.

b) $r(x)={\displaystyle \frac{1}{2}}\cdot x-{\displaystyle \frac{97}{16}}$

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