Question

Find the equation of the tangent line to f(x)=1/8 x^4-32/x^3 at X=2

Answer 1

Given the function

`f\mleft(x\mright)=(1)/(8)x^4-(32)/(x^3)`

at the point x=2

the derivative of the function is

`f^(\prime)(x)=(x^3)/(2)-(96)/(x^4)`

derivative represent the slope of the tangent line

since x=2

then

`f^(\prime)(2)=((2)^3)/(2)-(96)/((2)^4)`
`f^(\prime)(2)=4-6=-2`

then m=-22

the line equation is

`y-y1=m(x-x1)`

where

m=-2

x1=2

y1=...

`y1=(1)/(8)(2)^4-(32)/((2)^3)`
`y1=-2`

y1=-2

then

`y-y1=m(x-x1)`
`y-(-2)=-2(x-2)`

solving for y

`y+2=-2x+4`
`y=-2x+2`

tangent line y = -2x+2