Respuesta :

The derivate of a function f(x) is determinated as:

[tex]f^{\prime}(x)=\lim _{h\to0}\frac{f(x+h)-f(x)}{h}[/tex]

For the function

[tex]f(x)=x^2-2x[/tex]

First we have to determine de f ( x + h ) as follow:

[tex]f(x+h)=(x+h)^2-2(x+h)[/tex][tex]f(x+h)=x^2+2xh+h^2-2x-h^{}[/tex]

Then we calculate and simplify the coeficient in the first formula

[tex]\frac{f(x+h)-f(x)}{h}[/tex][tex]\frac{(x^2+2xh+h^2-2x-h^{})-(x^2-2x)}{h}[/tex][tex]\frac{x^2+2xh+h^2-2x-h^{}-x^2+2x}{h}=\frac{h^2+2xh-h}{h}[/tex]

[tex]\frac{h^2+2xh-h}{h}\text{ = }\frac{h(h+2x-1)}{h}=h+2x-1[/tex]So the derivate is:[tex]f^{\prime}(x)=\lim _{h\to0}\frac{f(x+h)-f(x)}{h}=\lim _{h\to0}h+2x-1[/tex][tex]f^{\prime}(x)=0+2x-1[/tex][tex]f^{\prime}(x)=2x-1[/tex]

------------------------------------------------------------------------------------The equation of the tangent:

First you need to know that the derivate of a function is equal to the slope (m) of the tangent of this function

And the equation to thist tangent in a specific point will be find using the next formula:

[tex]y-f(x_0)=m(x-x_0)_{}[/tex]

We have to calculate the slope in the point x =4 using the derivate:

[tex]m=2x-1[/tex][tex]m=2(4)-1=7[/tex]

In the point x=4

Calculate the value of f(x0) substituting in the function the given point x:

[tex]f(4)=4^2-2(4)\text{ = }8[/tex]

Knowing that we put the value of m and f(x0) in the equation of the tangent:

[tex]y-8=7(x-4)_{}[/tex][tex]y-8=7x-28[/tex][tex]y=7x-28+8[/tex]So the equation of the tangent in x= 4 is:[tex]y=7x-20[/tex]

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The normal line

The slope of the normal line is the opposite of the slope of theu tangent in an espesific point:

[tex]m_n=-\frac{1}{m_t}[/tex]

So in this situation is:

[tex]m_n=-\frac{1}{7}[/tex]

The equation of the normal line is given by the next formula:

[tex]y-f(x_0)=m_n(x-x_0)_{}[/tex]

Replacing the data we obtain:

[tex]y-8=-\frac{1}{7}(x-4)_{}[/tex][tex]y-8=-\frac{1}{7}x+\frac{4}{7}[/tex]So the equation of the normal line is:[tex]y=-\frac{1}{7}x+\frac{60}{7}[/tex]