I wish I could help but but I can't I'm dumb as a box of rocks
Answer:
Solution: (3, 2)
Step-by-step explanation:
It's easier to graph when the equation of the lines are in their slope-intercept form, y = mx + b.
<u>x - 3y = -3</u>
-3y = -x - 3
Divide both sides by -3:
y = 1/3x + 1
where slope = 1/3
y-intercept = 1
<u>x + y = 5</u>
Subtract x from both sides to isolate y:
x - x + y = - x + 5
y = -x + 5
where slope = -1
y-intercept, 5
I started by graphing the y-intercepts of each line. Then, I used the slope of each linear equation (rise over run) to plot the next points on the graph. In the attached screenshot of the graph, <u>x - 3y = -3</u> is the blue line, while <u>x + y = 5</u> is the green line. Their intersection occurs at point, (3, 2).
Therefore, the solution is (3, 2).
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14+-86=-72. That is the change in level if you needed it.
Hmm, the 2nd derivitve is good for finding concavity
let's find the max and min points
that is where the first derivitive is equal to 0
remember the difference quotient
so
f'(x)=(x^2-2x)/(x^2-2x+1)
find where it equals 0
set numerator equal to 0
0=x^2-2x
0=x(x-2)
0=x
0=x-2
2=x
so at 0 and 2 are the min and max
find if the signs go from negative to positive (min) or from positive to negative (max) at those points
f'(-1)>0
f'(1.5)<0
f'(3)>0
so at x=0, the sign go from positive to negative (local maximum)
at x=2, the sign go from negative to positive (local minimum)
we can take the 2nd derivitive to see the inflection points
f''(x)=2/((x-1)^3)
where does it equal 0?
it doesn't
so no inflection point
but, we can test it at x=0 and x=2
at x=0, we get f''(0)<0 so it is concave down. that means that x=0 being a max makes sense
at x=2, we get f''(2)>0 so it is concave up. that means that x=2 being a max make sense
local max is at x=0 (the point (0,0))
local min is at x=2 (the point (2,4))
