Heyy, please find the answers to the two questions below in the attachment. Thanks!
1 answer:
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The maxima of f(x) occur at its critical points, where f '(x) is zero or undefined. We're given f '(x) is continuous, so we only care about the first case. Looking at the plot, we see that f '(x) = 0 when x = -4, x = 0, and x = 5.
Notice that f '(x) ≥ 0 for all x in the interval [0, 5]. This means f(x) is strictly increasing, and so the absolute maximum of f(x) over [0, 5] occurs at x = 5.
By the fundamental theorem of calculus,
The definite integral corresponds to the area of a trapezoid with height 2 and "bases" of length 5 and 2, so
Answer: < -4/5, 3/5>
This is equivalent to writing < -0.8, 0.6 >
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Explanation:
Draw an xy grid and plot the point (-4,3) on it. Draw a segment from the origin to this point. Then draw a vertical line until reaching the x axis. See the diagram below.
We have a right triangle with legs of 4 and 3. The hypotenuse is through use of the pythagorean theorem.
We have a 3-4-5 right triangle.
Therefore, the vector is 5 units long. This is the magnitude of the vector.
Divide each component by the magnitude so that the resulting vector is a unit vector pointing in this same direction.
Therefore, we go from < -4, 3 > to < -4/5, 3/5 >
This is equivalent to < -0.8, 0.6 > since -4/5 = -0.8 and 3/5 = 0.6
Side note: Unit vectors are useful in computer graphics.
