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Given that

f(x) = x^{3} + x

**A. Domain: **

Clearly y is defined for all x in (-∞, ∞). so, domain of

y = (-∞, ∞)

**B. Intercept: **

Put x = 0 then y = 0

Put y = 0 then x (x^{2 } + 1) = 0 => x = 0, x^{2} = -1

this gives no real value.

So, the curve meets the axes only at (0, 0).

**C. Symmetry:**

Here,

f(-x) = (-x)^{3} + (-x) = -x^{3} – x = -f(x)

This means y is an odd function, so it is symmetrical about x-axis

**D. ****Asymptotes**

Here,

\(\lim_{x \to ∞} (y) = ∞\)

and

\(\lim_{x \to a} (y) = \lim_{x \to a} (x^3 + x) = a^3 + a \neq ∞\) for any finite value of a

This means y has no asymptotes

**E. Extreme:**

Here f(x) = 3x^{2} + 1

For the Critical Point, set

f'(x) = 0

\(\Rightarrow 3x^2 + 1 = 0\)

\(\Rightarrow x^2 = -\frac{1}{3} \Rightarrow x = no \enspace real \enspace value\)

This means y has no critical point. So,

Interval | Sign of f^{‘}(x) |
Nature of f(x) |

(-∞, ∞) | +ve | Increasing |

**F. Concavity:**

Here

f^{”}(x) = 6x

For Point of inflection, set

\(f^{”}(x) = 0 \Rightarrow x = 0\)

Interval | Sign of f^{”}(x) |
Nature of f(x) |

(-∞, 0) | -ve | Concave Down |

(0, ∞) | +ve | Concave Up |

**G. Summary**

Interval | Nature of f(x) | |

(-∞, 0) | Increasing | Increasing |

(0, ∞) | Concave down | Concave up |

With the help of these information, the sketch of the curve is as:

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