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Aleks [24]
3 years ago
9

Consider the function below, which has a relative minimum located at (-3 , -18) and a relative maximum located at (1/3, 14/17).

f(x) = -x^3 - 4x^2 + 3x Select all ordered pairs in the table which are located where the graph of f(x) is decreasing. (-1,-6) (2,-18) (0,0) (1,-2) (-3,-18) (-4,-12)
Mathematics
1 answer:
leonid [27]3 years ago
8 0
First of all, when I do all the math on this, I get the coordinates for the max point to be (1/3, 14/27).  But anyway, we need to find the derivative to see where those values fall in a table of intervals where the function is increasing or decreasing.  The first derivative of the function is f'(x)=-3x^2-8x+3.  Set the derivative equal to 0 and factor to find the critical numbers. 0=-3x^2-8x+3, so x = -3 and x = 1/3.  We set up a table of intervals using those critical numbers, test a value within each interval, and the resulting sign, positive or negative, tells us where the function is increasing or decreasing.  From there we will look at our points to determine which fall into the "decreasing" category.  Our intervals will be -∞<x<-3, -3<x<1/3, 1/3<x<∞.  In the first interval test -4. f'(-4)=-13; therefore, the function is decreasing on this interval.  In the second interval test 0. f'(0)=3; therefore, the function is increasing on this interval.  In the third interval test 1. f'(1)=-8; therefore, the function is decreasing on this interval.  In order to determine where our points in question fall, look to the x value.  The ones that fall into the "decreasing" category are (2, -18), (1, -2), and (-4, -12).  The point (-3, -18) is already a min value.
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See explanation

Step-by-step explanation:

1 step:

n=1, then

\sum \limits_{j=1}^1 2^j=2^1=2\\ \\2(2^1-1)=2(2-1)=2\cdot 1=2

So, for j=1 this statement is true

2 step:

Assume that for n=k the following statement is true

\sum \limits_{j=1}^k2^j=2(2^k-1)

3 step:

Check for n=k+1 whether the statement

\sum \limits_{j=1}^{k+1}2^j=2(2^{k+1}-1)

is true.

Start with the left side:

\sum \limits _{j=1}^{k+1}2^j=\sum \limits _{j=1}^k2^j+2^{k+1}\ \ (\ast)

According to the 2nd step,

\sum \limits_{j=1}^k2^j=2(2^k-1)

Substitute it into the \ast

\sum \limits _{j=1}^{k+1}2^j=\sum \limits _{j=1}^k2^j+2^{k+1}=2(2^k-1)+2^{k+1}=2^{k+1}-2+2^{k+1}=2\cdot 2^{k+1}-2=2^{k+2}-2=2(2^{k+1}-1)

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3 years ago
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Area: 180 units2 (units 2 is because since the are no specific unit given but every area should have a unit  of measurement)

Step-by-step explanation:

The area enclosed by the graph of the function, the horizontal axis, and vertical lines is the integral of the function between thos two points (x=2 and x=4)

So , let's solve the integral of f(x)

Area =\int\limits^2_4 3{x}^3 \, dx = 3*x^4/4+C

C=0

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Area= 3* (4^4)/4-3*(2^4)/4= 3*(4^4-2^4)/4=180 units 2

Goos luck!

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We don't know the numeric value of (x+8) since we don't know the value of x, so we leave it as is. 
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