#### To determine

**To sketch:**

the graph of the function y=ln(1+x3)

#### Answer

Please see the graph given below.

#### Explanation

**Steps in Curve tracing**

**A. Domain:** Given y=ln(1+x3)

Since it is defined for all positive values of x

Therefore, the domain is {x|x>0}

**B. Intercepts:**

y Intercept is obtained by taking x=0

Therefore, y=0

x intercept is obtained by taking y=0

Therefore, x=0

**C. Symmetry:**

Since f(−x)≠f(x), the curve is not symmetrical about both axis. The curve passes through origin

**D. Asymptotes:**

There are no horizontal asymptotes and vertical asymptotes

**E. Interval of increase and decrease**

y=ln(1+x3)y′=3x21+x3

Since numerator is positive, the sign of y′ depends on the denominator

The function is increasing when x is positive, therefore f(x) is increasing in (0,∞) and it is decreasing in (−∞,0)

**F. Local maxima and minima**

Maxima and minima are obtained by the first derivative test,

i.e., y′=3x21+x3=0⇒3x2=0⇒x=0

f(0)=0 is the local minima

**G. Concavity and inflexion points**

y′=3x21+x3y″=(1+x3).6x−3x2.3x2(1+x3)2=6x(1+x3)2y″=0⇒x=0

The function is concave up in (0, 1) and concave down in (−∞,−1) and (1,∞)

(0,0) is the inflexion point

**H.** Using this information we sketch the curve

**Final statement:** The sketch of the curve y=ln(1+x3) is given by