Question

Find f'(x) when f(x) = (x^2) - 2x. find the equation of the tangent line and the normal line at x=4

Answer 1

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

`f^(\prime)(x)=\lim _(h\to0)(f(x+h)-f(x))/(h)`

For the function

`f(x)=x^2-2x`

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

`f(x+h)=(x+h)^2-2(x+h)`
`f(x+h)=x^2+2xh+h^2-2x-h^{}`

Then we calculate and simplify the coeficient in the first formula

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

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

------------------------------------------------------------------------------------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:

`y-f(x_0)=m(x-x_0)_{}`

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

`m=2x-1`
`m=2(4)-1=7`

In the point x=4

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

`f(4)=4^2-2(4)\text{ = }8`

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

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

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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:

`m_n=-(1)/(m_t)`

So in this situation is:

`m_n=-(1)/(7)`

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

`y-f(x_0)=m_n(x-x_0)_{}`

Replacing the data we obtain:

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