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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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