Answer:
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Step-by-step explanation:
1. Find the equation of the line AB. For reference, the answer is y=(-2/3)x+2.
2. Derive a formula for the area of the shaded rectange. It is A=xy (where x is the length and y is the height).
3. Replace "y" in A=xy with the formula for y: y= (-2/3)x+2:
A=x[(-2/3)x+2] This is a formula for Area A in terms of x only.
4. Since we want to maximize the shaded area, we take the derivative with respect to x of A=x[(-2/3)x+2] , or, equivalently, A=(-2/3)x^2 + 2x.
This results in (dA/dx) = (-4/3)x + 2.
5. Set this result = to 0 and solve for the critical value:
(dA/dx) = (-4/3)x + 2=0, or (4/3)x=2 This results in x=(3/4)(2)=3/2
6. Verify that this critical value x=3/2 does indeed maximize the area function.
7. Determine the area of the shaded rectangle for x=3/2, using the previously-derived formula A=(-2/3)x^2 + 2x.
The result is the max. area of the shaded rectangle.
Answer:
C) As x approaches positive infinity, f(x) approaches positive infinity
Step-by-step explanation:
- The domain is NOT all real numbers as x is either smaller than or bigger than 0, and smaller than or bigger than 2. So x ≠ 0 and x ≠ 2.
- This implies that there are asymptotes at x=0 and x=2.
Therefore, the function is NOT continuous.
- The function is NOT increasing over its entire domain as
f(x) = -x² -4x + 1 is decreasing for its given domain of 0<x<2
Answer:
Step-by-step explanation:
If the graph crosses the x-axis and appears almost linear at the intercept, it is a single zero. If the graph touches the x-axis and bounces off of the axis, it is a zero with even multiplicity. If the graph crosses the x-axis at a zero, it is a zero with odd multiplicity.