# Thread: calculus inflection, increasing, decreasing

1. ## calculus inflection, increasing, decreasing

1.On what interval the function f(x)=x/(x^2+1) increasing?

(a)[-1,1](b)[-inf,1] (c) [-1,2] (d) [-inf, 2]

2. find the interval on which the function f(x)=x+cosx 0<=x<=2pi is increasing
(a)(pi/3,5pi/3)
(b)5pi/3,2pi)
(c) (pi/2, 3pi/2)
(d)(0,2pi,)
(e)none
3.
How many pointsof inflection does f(x) =x^3e^(-x) have?
(a)0
(b)1
(c)2
(d)3
(e) none
4. consider the function f(x)=x^2 in interval [0,1/2].According to mean value theorem thre must be a number c in (0,1/2) such that f '(c)is equal to particular value d. what is d?
(a) 3/2
(b)1
(c) 1/2
(d)2
(e) none

2. Originally Posted by bobby77
4. consider the function f(x)=x^2 in interval [0,1/2].According to mean value theorem thre must be a number c in (0,1/2) such that f '(c)is equal to particular value d. what is d?
(a) 3/2
(b)1
(c) 1/2
(d)2
(e) none
$\displaystyle f'(c)=\frac{f(1/2)-f(0)}{1/2-0}$
Thus,
$\displaystyle f'(c)=\frac{1/4}{1/2}=\frac{1}{2}$
Thus that number is $\displaystyle d=1/2$
How many pointsof inflection does f(x) =x^3e^(-x) have?
Then,
$\displaystyle y'=3x^2e^{-x}-x^3e^{-x}$
$\displaystyle y''=6xe^{-x}-3x^2e^{-x}-3x^2e^{-x}+x^3e^{-x}$
Thus,
$\displaystyle 6xe^{-x}-6x^2e^{-x}+x^3e^{-x}=0$
Thus,
$\displaystyle 6x-6x^2+x^3=0$
Thus,
$\displaystyle x(x^2-6x+6)=0$
$\displaystyle 6^2-4\cdot 6>0$ so it has two solution (non-trivial) thus, in total 3 solutions.
Corresponding to 3 inflections points.

3. Hello, Bobby!

3. How many points of inflection does $\displaystyle f(x) = x^3e^{-x}$ have?

. . . $\displaystyle (a)\;0\qquad(b)\;1\qquad(c)\;2\qquad(d)\;3\qquad(e )\text{ none}$

Points of inflection occur where $\displaystyle f''(x) = 0$

We have: .$\displaystyle f'(x)\;=\;-x^3e^{-x} + 3x^2e^{-x}$
Then: .$\displaystyle f''(x)\;=\;x^3e^{-x} - 3x^2e^{-x} - 3x^2e^{-x} + 6xe^{-x} \;=\;x^3e^{-x} - 6x^2e^{-x} + 6xe^{-x}$

We have: .$\displaystyle xe^{-x}(x^2 - 6x + 6) \:=\:0$

Hence, the roots are: .$\displaystyle x\,=\,0$
. . and: .$\displaystyle x^2 - 6x + 6 \:=\:0\quad\Rightarrow\quad x \,=\,3\pm\sqrt{3}$

Therefore, there are three inflection points . . . answer (d).