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Find the minimum value of \frac{x^2}{x - 1 + x^3} for x > 1    

I used Desmos to plot y = (x²) / (x – 1 + x³)    

As x increases past (1,1), the value of y decreases asymptotically,    

which means y approaces zero (the x-axis) but it never gets there.    

So there is no minimum value.  We say "it approaches zero".    

_.    

The range is [4,5].

The domain is [0,inf).

To find the area between the curves y = 6x² - 18x and y = -6x from x = 1 to x = 3, we need to integrate the absolute difference of the two functions over the given interval.

Find the difference between the functions: (6x² - 18x) - (-6x) = 6x² - 18x + 6x = 6x² - 12x

Determine where the functions intersect (if they do within the interval): Set 6x² - 18x = -6x 6x² - 12x = 0 6x(x - 2) = 0 x = 0 or x = 2

Since x = 2 is within our interval [1, 3], we need to consider the intervals [1, 2] and [2, 3] separately.

Integrate the absolute difference of the functions over the intervals:

Interval [1, 2]: In this interval, -6x is greater than or equal to 6x^2 - 18x.

So we integrate: ∫(from 1 to 2) |-6x - (6x² - 18x)| dx = ∫(from 1 to 2) |-6x² + 12x| dx = ∫(from 1 to 2) (-6x² + 12x) dx = [-2x³ + 6x²] (from 1 to 2) = (-16 + 24) - (-2 + 6) = 8 - 4 = 4

Interval [2, 3]: In this interval, 6x² - 18x is greater than or equal to -6x.

So we integrate: ∫(from 2 to 3) |(6x² - 18x) - (-6x)| dx = ∫(from 2 to 3) |6x² - 12x| dx = ∫(from 2 to 3) (6x² - 12x) dx = [2x³ - 6x²] (from 2 to 3) = (54 - 54) - (16 - 24) = 0 - (-8) = 8

Add the absolute values of the integrals: Total area = |4| + |8| = 4 + 8 = 12

Therefore, the total area of the region between the curves y = 6x² - 18x and y = -6x from x = 1 to x = 3 is 4.

From the second sentence, we know that triangle ABC is a right isosceles triange and that B is a right angle. As BD bisects angle ABC, angle ABD is also 45 degrees, meaning triangle ABD is another right isosceles triangle. From the angle bisector theorem, we can figure out that AD=BD=6, so therefore, the area of the triangle is 6*6*1/2=18.

Feel free to tell me if I did anything wrong! :D

The answer is (b,c,d) = (-2,8,15).

The answer is (b,c) = (5,-15).

The roots of f(x) are -1, -8, and -10.

The real numbers x and y satisfy    
x^2 + y^2 - 8x + 6y + 23 = 0.    
Find the largest possible value of x + y.      

Given                               x² + y² – 8x + 6y + 23  =  0     

Group together               (x² – 8x     )  +  (y² + 6y     )  + 23  =  0    

                                       (x² – 8x     )  +  (y² + 6y     )           =  – 23   

Complete both squares and add the same    

amount on the right as you add on the left.    

                                         (x² – 8x + 16) + (y² + 6y + 9)  =  – 23  + 16  + 9     

                                         (x – 4)²  +  (y + 3)²  =  2    

This draws a circle centered at (+4, –3) with a radius the sqrt(2)     

We don't care about the center, all we care about is the radius.    

By formula, the largest x+y of a circle is its radius times the sqrt(2).    

So, the largest x+y is the sqrt(2) times the sqrt(2)         x + y = sqrt(2) • sqrt(2)    

                                                                                       largest x + y = 2    

_.    

The possible values of m are -5, 1, 2, and 13.

\( (2 + \sqrt{5})(137) = a + b \sqrt{5} \)

Simplify as

274  + 137sqrt (5)

a  = 274 =  137*2        b = 137

a^2  - 5b^2   =

(a  - sqrt (5) b)  ( a +  sqrt(5) b)   =

(137 * 2  - 137 sqrt (5)) ( 137 *2   137 sqrt (5))  

137(2 - sqrt 5) * 137 (2 + sqrt (5) ) =

137^2 *  ( 2 - sqrt 5) (2 + sqrt 5)  

137^2 ( 4 - 5)  =

137^2 (-1)   

-137^2    =   

-18769

\(What\ is\ the\ equation\ of\ the\ line\ of\ symmetry\ of\ the\ parabola\ y = f(x)?\)

\(f(x) = 2x^2 - 13x + 20 - 5x^2 + 19x + 7\\ f(x)= -3x^2+6x+27\\ x^2-2x-9=0\\ x_{1,2}=1\pm\sqrt{1^2+9}\\ x\in \{(1-\sqrt{10}),(1+\sqrt{10})\}\\ \dfrac{1-\sqrt{10}+1+\sqrt{10}}{2}=1\)

\({\color{blue}The\ equation\ of\ the\ line\ of\ symmetry}\ of\ the\ parabola\ y=f(x)\ \color{blue}is\ x=1 \)

 !

log (y) + log (2)  = log (a^(-x) + a^x)

log (2y) =  log ( 1/a^x + a^x)

log (2y)  = log  [ (1+ a^(2x)) / a^x]

2y  =  ( 1 + a^(2x) )  / a^x 

2y (a^x)  = 1 + (a^x)^2        let a^x =  m

2y (m)  = 1  + m^2

m^2 - (2y)m  + 1  =  0

m^2 - (2y)m + y^2  =  - 1 + y^2

(m - y)^2  =  y^2 -1         take the positive root

m - y =  sqrt [y^2 -1 ] 

m =  sqrt [ y^2 - 1  ] + y

a^x  = sqrt [ y^2 - 1 ] + y

log a^x = log [ sqrt ( y^2  -1) + y ]

x log a  = log [ sqrt ( y^2 - 1)  + y ] 

x =  log [ sqrt (  y^2 - 1) +  y ]

      ____________________         ,   { y  ≥ 1 }

               log a

The positive integers n that work are -10, 5, 23, and 45.

The range is [-5, 1/4].

Simplify  as

4y^2  -11y    - 4x^2  -8x    =  15         complete the square on x, y

4 (y^2 - (11/4)y  + 121/64)  -  4(x^2 + 2x + 1)  =  15  + 121/64 - 4   

4( y - 11/8)^2  -  4(x + 1)^2  =  297/16

(y -11/8)^2   -    ( x + 1)^2 =  1

__________     ________

  297/64            297/64

Vertices =  (-1, 11/8  +/- sqrt (297) / 8 )

6x^2 -18x + 17  + 3y^2 + 6y  + 11  = 28

x + y = 20

y  = 20 - x

So

6x^2 - 18x + 17 + 3(20 -x)^2 + 6(20 -x) + 11  = 28

6x^2  -18x + 3(x^2 - 40x + 400)  + 120 -6x + 28  = 28

6x^2 - 18x + 3x^2 -120x + 1200 + 120 -6x  =  0

9x^2 - 144x + 1320  =  0

No real solutions for this since the discriminant   144^2  - 36*1320 = -26784  is negative

-4x^2 + y^2 -2y = -3y^2 + 8x + 9y + 15

4y^2 - 11y   -  4x^2  - 8x  =  15     complete the square on  x, y

4(y^2 -(11/4)y + 121/64) - 4(x^2 + 2x + 1)   =  15 + 121/16 - 4

4( y - 11/8)^2  -  4(x + 1)^2   =  297/16

Center  = (-1, 11/8)       

Simplify as

6x^3  - 34x^2  - 60x - 225  =  0 

No integer roots

Only real root ≈  7.62

https://www.wolframalpha.com/input?i=x%5E3+-+14x%5E2+-+67x+-+90+%2B+5x%5E3+-+20x%5E2+%2B+7x+-+135+%3D0

Simplify as

-33x  -8y^2  + 11y  + 29  = 0

33x + 8y^2 - 11y  - 29  = 0

33x  =  -8y^2 + 11y  + 29

x =  (-8y^2 + 11y  + 29) / 33

x to 1st power, y to second power =   negative in front of y^2  = parabola  opening to the left

Let  p(x)  = ax^3 + bx^2 + cx + d

If p(0) = 0, then  d = 0

So   we have the equations

a(1)^3 + b(1)^2 + c(1)  = 1

a(2)^3 + b(2)^2 + c(2)  = 2

a(3)^3 + b(3)^2 + c(3)   =3       simplify these

a + b +  c  = 1                    (1)

8a +4b + 2c  = 2                (2)

27a + 9b + 3c  = 3             (3)

Multiply (1) by -2  add to (2)   and multiply 1 by -3  and add to (3)

6a  + 2b =  0    →  3a + b =  0   →  -3a - b = 0   ( 4)

24a + 6b  = 0   →  4a + b = 0  (5)

Add (4) , (5)     a = 0      and  so       b = 0

No such cubic polynomial exists

Let  g(x)  = x^3  + bx^2  + cx + d

                        x  

x^2  - x  - 6   [  x^3  + bx^2 + cx  + d ] 

                        x^3  -  x^2    - 6x

                      __________________

b + 1  = 0       c + 6  = 2       d = 7

b = -1              c = -4 

So g(x)  =  x^3 -x^2 -4x + 7

g(8) =  8^3  - 8^2 - 4(8) + 7   =  423