Sketch the curve y = x^3 + x

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Suresh Chand · Accepted Answer

Solution: Given that f(x) = x3 + 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 (x2  + 1) = 0 => x = 0, x2 = -1 this gives no real value. So, the curve meets the axes only at (0, 0). C. Symmetry: Here, f(-x) = (-x)3 + (-x) = -x3 - 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) = 3x2 + 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:

Sketch the curve y = x^3 + x

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Sketch the curve y = x^3 + x

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