Welcome :: Homework Help and Answers :: Mathskey.com

Recent Visits

    
Welcome to Mathskey.com Question & Answers Community. Ask any math/science homework question and receive answers from other members of the community.

13,459 questions

17,854 answers

1,446 comments

790,617 users

label the relatvie extrema, point of inflection, asymptote and graph

0 votes
f(x)=4x^3-x^4

 

f(x)=(1)/(1+x^2)
asked Oct 16, 2014 in CALCULUS by anonymous

2 Answers

0 votes

The function is f(x) = 4x3 - x4.

f,(x) = 12x2 - 4x3.

f,,(x) = 24x - 12x2.

To find the critical or key numbers, to make the first derivative equal to zero or f ' (x) does not exist.

f,(x) = 12x2 - 4x3 = 4x2 (3 - x) = 0.

The key numbers are x = 0 and x = 3.

(b).

TEST FOR INCREASING AND DECREASING FUNCTIONS :

If f,(x) > 0 (Positive) for all x in (a, b), then f(x) is increasing on [a, b].

If f,(x) < 0 (Negative) for all x in (a, b), then f(x) is decreasing on [a, b].

Test intervals    x - Value                    Sign of f,(x)                             Conclusion

(- ∞, 0)              x = - 1         12(-1)2 - 4(-1)3 = 12 + 4 = 16 > 0        Increasing

(0, 3)                 x = 1         12(1)2 - 4(1)3 = 12 - 4 = 8 > 0               Increasing

(3, ∞)                x = 4          12(4)2 - 4(4)3 = 192 - 256 = - 64 < 0    Decreasing

So, f(x) is increasing on the interval (- ∞, 3) and decreasing on the interval (3, ∞).

Find Relative Extrema :
The First Derivative Test :
Let be a differential function with f(c) = 0 then

  • If f ' (x) changes from positive to negative, f has a relative (local) maximum at c.
  • If f ' (x) changes from negative to positive, f has a relative (local) minimum at c.

The function f(x) has relative maximum at x = 3, because f'(x) changes from positive to negative around 3.

f(3) = 4(3)3 - 34 =108 - 81 = 27.

The relative maximum point (3, 27).

Points of Inflection : If (c, f(c)) is a point of inflection of the graph of f(x), then either f " (c) = 0 or f " (x) does not exist.

f,,(x) = 24x - 12x2 = 12x(2 - x) = 0.

The key numbers are x = 0 and x = 2.

If x = 0 then, f(0) = 4(0)3 - (0)= 0.

If x = 2 then, f(2) = 4(2)3 - (2)= 32 - 16 = 16.

The Inflection points are (0, 0) and (2, 16).

Graph :

answered Oct 16, 2014 by casacop Expert
edited Oct 16, 2014 by bradely
0 votes

(2).

The function is f(x) = 1/(1 + x2).

To find the vertical asymptotes, denominator function is zero.

1 + x2 = 0

x2 = - 1

x = ± i.

Since the imaginary numbers are negligible, there is no vertical asymptotes.

Since degree of numerator (0) < degree of denominator (2), the horizontal asymptote y = 0.

The function is .

Differentiate with respect to x.

Again differentiate with respect to x.

To find the critical numbers, image or image does not exist.

image

The function image does not exist when image.

The imaginary numbers are negligible, so the critical number is x = 0.

Apply second derivative test:

image

image

The relative maximum point (0, 1).

To find the points of inflection, image or image does not exist.

image

The function image does not exist when image

The imaginary numbers are negligible.

image

The points of inflections are (1, 1/2) and (-1, 1/2).

Graph :

 

answered Oct 17, 2014 by casacop Expert

Related questions

asked Nov 5, 2014 in PRECALCULUS by anonymous
asked Nov 18, 2014 in PRECALCULUS by anonymous
...