# Sketch the curve y = x^3 + x

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 (x + 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: