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

Sagot :

Given the function

[tex]f\mleft(x\mright)=\frac{1}{8}x^4-\frac{32}{x^3}[/tex]

at the point x=2

the derivative of the function is

[tex]f^{\prime}(x)=\frac{x^3}{2}-\frac{96}{x^4}[/tex]

derivative represent the slope of the tangent line

since x=2

then

[tex]f^{\prime}(2)=\frac{(2)^3}{2}-\frac{96}{(2)^4}[/tex][tex]f^{\prime}(2)=4-6=-2[/tex]

then m=-22

the line equation is

[tex]y-y1=m(x-x1)[/tex]

where

m=-2

x1=2

y1=...

[tex]y1=\frac{1}{8}(2)^4-\frac{32}{(2)^3}[/tex][tex]y1=-2[/tex]

y1=-2

then

[tex]y-y1=m(x-x1)[/tex][tex]y-(-2)=-2(x-2)[/tex]

solving for y

[tex]y+2=-2x+4[/tex][tex]y=-2x+2[/tex]

tangent line y = -2x+2

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