Answer:
Step-by-step explanation:
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now, we get critical points from zeroing out the derivative, and also from zeroing out the denominator, but those at the denominator are critical points where the function is not differentiable, namely a sharp spike or cusp or an asymptote.
so, from zeroing out the derivative we get no critical points there, from the denominator we get x = 8, but can't use it because f(x) is undefined.
therefore, we settle for the endpoints, 4 and 6,
f(4) =3 and f(6) = 7
doing a first-derivative test, we see the slope just goes up at both points and in between, but the highest is f(6), so the absolute maximum is there, while we can take say f(4) as the only minimum and therefore the absolute minumum as well.
Answer:
17 .09 is the decimal equivalent to 188/11 .
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Answer:
We want a line that passes through the point (4, 7/2)
and we have no other information of this line, so we can not fully find it, but we can find a general line.
We know that a line can be written as:
y = a*x + b.
Now we want that, when x = 4, we must have y = 7/2.
7/2 = a*4 + b
b = -a*4 + 7/2
Then we can write this line as:
y = a*x - a*4 + 7/2.
Where a can take any value, and it is the slope of our line.