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Fittoniya [83]
3 years ago
6

Locate the absolute extrema of the function on the closed interval: y = 3x^(2/3) - 2x, [-1, 1]

Mathematics
1 answer:
Alecsey [184]3 years ago
7 0
To get the extrema, derive the function.
You get y' = 2x^-1/3 - 2.
Set this equal to zero, and you get x=0 as the location of a critical point.
Since you are on a closed interval [-1, 1], those points can also have an extrema.
Your min is right, but the max isn't at (1,1). At x=-1, you get y=5 (y = 3(-1)^2/3 -2(-1); (-1)^2/3 = 1, not -1).
Thus, the maximum is at (-1, 5).
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Parabolas y=−2x^2 and y=2x^2 +k intersect at points A and B that are in the third and the fourth quadrants respectively. Find k
OverLord2011 [107]

Answer:

-25  

Step-by-step explanation:

(1)   y = -2x²

(2) y =   2x² + k     Subtract (1) from (2)

    0 =  4x² + k      Subtract 4x² from each side

    k = -4x²

The parabolas are <em>symmetrical about the y-axis.</em>

Segment AB = 5, so x = +2.5 and x = +2.5.

k = -4(±2.5)² = -4 × 6.25 = -25

8 0
3 years ago
Please help me with mathh !!!
larisa86 [58]
Okay I can help you with your math!!
8 0
2 years ago
Read 2 more answers
Ms. Maxwell bought a purse with 1/4 of her money. The purse cost 37.00. How much money did Ms. Maxwell have left after she bough
bixtya [17]

Answer:111

the purse was 1/4 so. we have to do

37.00x4 we get 148 then. we do 148-37 to get a total of 111 (:

Step-by-step explanation:

7 0
2 years ago
What are the roots of this equation? x2-4x+9=0
zysi [14]

Considering the definition of zeros of a function, the zeros of the quadratic function f(x) = x² + 4x +9 do not exist.

<h3>Zeros of a function</h3>

The points where a polynomial function crosses the axis of the independent term (x) represent the so-called zeros of the function.

In summary, the roots or zeros of the quadratic function are those values ​​of x for which the expression is equal to 0. Graphically, the roots correspond to the abscissa of the points where the parabola intersects the x-axis.

In a quadratic function that has the form:

f(x)= ax² + bx + c

the zeros or roots are calculated by:

x1,x2=\frac{-b+-\sqrt{b^{2}-4ac } }{2a}

<h3>This case</h3>

The quadratic function is f(x) = x² + 4x +9

Being:

  • a= 1
  • b= 4
  • c= 9

the zeros or roots are calculated as:

x1=\frac{-4+\sqrt{4^{2}-4x1x9 } }{2x1}

x1=\frac{-4+\sqrt{16-36 } }{2x1}

x1=\frac{-4+\sqrt{-20 } }{2x1}

and

x2=\frac{-4-\sqrt{4^{2}-4x1x9 } }{2x1}

x2=\frac{-4-\sqrt{16-36 } }{2x1}

x2=\frac{-4-\sqrt{-20} }{2x1}

If the content of the root is negative, the root will have no solution within the set of real numbers. Then \sqrt{-20} has no solution.

Finally, the zeros of the quadratic function f(x) = x² + 4x +9 do not exist.

Learn more about the zeros of a quadratic function:

brainly.com/question/842305

brainly.com/question/14477557

#SPJ1

6 0
1 year ago
I just need help with #16 … can someone please help me
laiz [17]

Answer: it goes up like 10,20,30,40

Step-by-step explanation:

7 0
3 years ago
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