The function fx - 4- x2-4x - x3 is

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Geometrical Explanation of Intermediate Value Theorem

The function ...

Question

The function $f (x) = \frac{4 - x^{2}}{4 x - x^{3}}$ is

discontinuous at only one point

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discontinuous exactly at two points

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discontinuous exactly at three points

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None of these

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f (x) = \frac{4 - x^{2}}{4 x - x^{3}}

f (x) = | x^{2} - 2 x - 3 | \cdot e^{| 9 x^{2} - 12 x + 4 |}

C_{1} : y^{2} = 4 x, C_{2} : x^{2} + y^{2} - 6 x + 1 = 0.

Let f : [- \frac{1}{2}, 2] \to R and g : [- \frac{1}{2}, 2] \to R

f (x) = [x^{2} - 3]

g (x) = | x | f (x) + | 4 x - 7 | f (x),

[y]

y

y \in R .

g (x) = x (1 - x^{2})

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