Application of DerivativeMedium
Question
Consider the curve f(x) = x1/3, then -
Options
A.the equation of tangent at (0, 0) is x = 0
B.the equation of normal at (0, 0) is y = 0
C.normal to the curve does not exist at (0, 0)
D.f(x) and its inverse meet at exactly 3 points.
Solution
f′(x) = 
f′(0) → ∞ tangent is vertical at x = 0
Equation of tangent at (0, 0) is x = 0
Equation of normal is y = 0
f(x) = f1(x)
x1/3 = x3 ⇒ x9 = x
⇒ x = 0;1; -1
f′(0) → ∞ tangent is vertical at x = 0
Equation of tangent at (0, 0) is x = 0
Equation of normal is y = 0
f(x) = f1(x)
x1/3 = x3 ⇒ x9 = x
⇒ x = 0;1; -1
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