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2 years ago

Find the intervals on which the function is increasing or decreasing.

Mathematics

1 answer:

Assoli18 [71]2 years ago

6 0

To solve this, you have to know that the first derivative of a function is its slope. When an interval is increasing, it has a positive slope. Thus, we are trying to solve for when the first derivative of a function is positive/negative.

f(x)=2x^3+6x^2-18x+2
f'(x)=6x^2+12x-18
f'(x)=6(x^2+2x-3)
f'(x)=6(x+3)(x-1)
So the zeroes of f'(x) are at x=1, x=-3
Because there is no multiplicity, when the function passes a zero, he y value is changing signs.
Since f'(0)=-18, intervals -3<x<1 is decreasing(because -3<0<1)
Thus, every other portion of the graph is increasing.
Therefore, you get:

Increasing: (negative infinite, -3), (1, infinite)
Decreasing:(-3,1)

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Which value, when placed in the box, would result in a system of equations with infinitly many solutions? Y=2x - 5 2y-4x=

Nezavi [6.7K]

Given:

The system of equations is

$y=2x-5$

$2y-4x=?$

To find:

The missing value for which the given system of equations have infinitely many solutions.

Solution:

Let the missing value be k.

We have,

$y=2x-5$

$2y-4x=k$

Taking all the terms on the left side, the given equations can be rewritten as

$-2x+y+5=0$

$-4x+2y-k=0$

The system of equations $a_1x+b_1y+c_1=0$ and $a_2x+b_2y+c_2=0$ have infinitely many solutions if

$\dfrac{a_1}{a_2}=\dfrac{b_1}{b_2}=\dfrac{c_1}{c_2}$

We have,

$a_1=-2,b_1=1,c_1=5$

$a_2=-4,b_2=2,c_2=-k$

Now,

$\dfrac{-2}{-4}=\dfrac{1}{2}=\dfrac{5}{-k}$

$\dfrac{1}{2}=\dfrac{1}{2}=\dfrac{5}{-k}$

$\dfrac{1}{2}=\dfrac{5}{-k}$

On cross multiplication, we get

$-k=10$

$k=-10$

Therefore, the missing value is -10.

6 0

2 years ago

BRAINLIEST BRAINLIEST BRAINLIEST BRAINLIEST<br> answer NUMBER 3 please answer quickly thank you

kherson [118]

I can’t answer anything from my side so I will guide you the best I can! Measure the highlighted lines on the image. Then record the numbers into the correct areas(length,width,height). Then multiply all of the numbers above. That’s how you answer question 3. I’m sorry that I can’t do much more on my side.