Question

What is the equation of the line tangent to `f(x)=sqrt(5+x)` at `x=-1`?

Answer 1

Equation of the line is
`x-4y=-9" "`the tangent line

Explanation:

From the given `f(x)=sqrt(5+x)` at `x=-1`

We need to solve for the slope at the given point. Obtain the first derivative `f' (x)`, then find slope `m=f' (-1)`

The first derivative
`d/dx(f(x))=f' (x)=d/dx(sqrt(5+x))`

`f' (x)=1/(2sqrt(5+x))*d/dx(5+x)`

`f' (x)=1/(2sqrt(5+x))*(d/dx(5)+d/dx(x))`

`f' (x)=1/(2sqrt(5+x))*(0+1)`

`f' (x)=1/(2sqrt(5+x))`

Determine now the slope `m=f' (-1)`

`m=f' (-1)=1/(2sqrt(5+(-1)))=1/(2sqrt4)=1/4`

We now have the slope. We need to find the ordinate `y=f(-1)` using `x=-1`
`f(x)=sqrt(5+x)`
`f(-1)=sqrt(5+(-1))`

`f(-1)=sqrt(4)`
`f(-1)=2`

the required point `(-1, 2)`

We can now solve for the equation of the tangent line using Point-Slope Form

`y-y_1=m(x-x_1)`

`y-2=1/4(x--1)`
`4(y-2)=x+1`
`4y-8=x+1`

`x-4y=-9" "`the tangent line
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Kindly see the graphs of `f(x)=sqrt(5+x)` and the tangent line `x-4y=-9" "` at `(-1, 2)`
graph{((y-sqrt(5+x))(x-4y+9))=0[-7,5,-3,3]}

God bless....I hope the explanation is useful.