A) y = 0
B) y = 1
C) y = 2
D) y = 3
Correct Answer: D
Solution :
\[y=x+\frac{4}{{{x}^{2}}}\] \[\frac{dy}{dx}=1-\frac{8}{{{x}^{3}}}=0\] \[1=\frac{8}{{{x}^{3}}}\] \[{{x}^{3}}=8\] \[x=2\] at \[x=2,\] \[y=x+\frac{4}{{{x}^{2}}}\] \[=2+\frac{4}{4}=3\] tangent \[y3=0(x2)\] \[y=3\]You need to login to perform this action.
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