Question

for the following functions, determine the:
(i) intervals where ( f ) is increasing or decreasing,
(ii) local maxima and minima of ( f ),
(iii) intervals where ( f ) is concave up and concave down,
(iv) inflection points of ( f ).
then roughly sketch the graph.

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

Explanation:

Step1: Find the first derivative

The first derivative \(f'(x)\) of \(f(x)=x + x^{2}-x^{3}\) is \(f'(x)=1 + 2x-3x^{2}\).
Set \(f'(x)=0\), so \(3x^{2}-2x - 1=0\).
Using the quadratic formula \(x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\) with \(a = 3\), \(b=-2\), \(c=-1\), we get \(x=\frac{2\pm\sqrt{4 + 12}}{6}=\frac{2\pm4}{6}\).
The critical points are \(x = 1\) and \(x=-\frac{1}{3}\).

Step2: Determine increasing/decreasing intervals

Test intervals:

• For \(x<-\frac{1}{3}\), let \(x=-1\), \(f'(-1)=1-2 - 3=-4<0\), so \(f(x)\) is decreasing on \((-\infty,-\frac{1}{3})\).
• For \(-\frac{1}{3}
• For \(x>1\), let \(x = 2\), \(f'(2)=1 + 4-12=-7<0\), so \(f(x)\) is decreasing on \((1,\infty)\).

Step3: Find local maxima and minima

Using the first - derivative test:

• At \(x =-\frac{1}{3}\), since \(f(x)\) changes from decreasing to increasing, \(f(-\frac{1}{3})=-\frac{1}{3}+\frac{1}{9}+\frac{1}{27}=-\frac{9 + 3-1}{27}=-\frac{5}{27}\) is a local minimum.
• At \(x = 1\), since \(f(x)\) changes from increasing to decreasing, \(f(1)=1 + 1-1 = 1\) is a local maximum.

Step4: Find the second derivative

The second derivative \(f''(x)\) of \(f(x)\) is \(f''(x)=2-6x\).
Set \(f''(x)=0\), then \(2-6x=0\), \(x=\frac{1}{3}\).

Step5: Determine concavity

• For \(x<\frac{1}{3}\), let \(x = 0\), \(f''(0)=2>0\), so \(f(x)\) is concave up on \((-\infty,\frac{1}{3})\).
• For \(x>\frac{1}{3}\), let \(x = 1\), \(f''(1)=2 - 6=-4<0\), so \(f(x)\) is concave down on \((\frac{1}{3},\infty)\).

Step6: Find inflection point

When \(x=\frac{1}{3}\), \(f(\frac{1}{3})=\frac{1}{3}+\frac{1}{9}-\frac{1}{27}=\frac{9 + 3-1}{27}=\frac{11}{27}\). So the inflection point is \((\frac{1}{3},\frac{11}{27})\).

Answer:

(i) \(f(x)\) is decreasing on \((-\infty,-\frac{1}{3})\cup(1,\infty)\) and increasing on \((-\frac{1}{3},1)\).
(ii) Local minimum at \((-\frac{1}{3},-\frac{5}{27})\) and local maximum at \((1,1)\).
(iii) Concave up on \((-\infty,\frac{1}{3})\) and concave down on \((\frac{1}{3},\infty)\).
(iv) Inflection point at \((\frac{1}{3},\frac{11}{27})\).