f(x) = 6 + 2x2 − x4 (a) Find the interval of increase. (Enter your answer using interval notation.) Find the interval of decreas
e. (Enter your answer using interval notation.) (b) Find the local minimum value(s). (Enter your answers as a ,-separated list. If an answer does not exist, enter DNE.) Find the local maximum value(s). (Enter your answers as a ,-separated list. If an answer does not exist, enter DNE.) (c) Find the inflection points. (x, y) = (smaller x-value) (x, y) = (larger x-value) Find the interval where the graph is concave upward. (Enter your answer using interval notation.) Find the interval where the graph is concave downward. (Enter your answer using interval notation.
1 answer:
Answer:
(a) increasing: (-∞, -1) U (0, 1); decreasing: (-1, 0) U (1, ∞)
(b) local minimum: (0, 6); local maxima (also global maxima): (-1, 7), (1, 7)
(c) points of inflection: (-(√3)/3, 6 5/9), ((√3)/3, 6 5/9)
(d) concave upward: (-(√3)/3, (√3)/3); concave downward: (-∞, -(√3)/3) U ((√3)/3, ∞)
Step-by-step explanation:
Most of the answers can be learned from the first derivative:
f'(x) = 4x -4x^3 = 4x(1-x^2)
This has zeros at x={-1, 0, 1}.
For points of infection, we also need to know the points where the second derivative is zero:
f''(x) = 4(-3x^2 +1)
This has zeros at x = ±1/√3
The attached plot shows the function f(x) (red) and the first derivative f'(x) (dashed blue), along with labels on most of the points of interest.
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(a) The function is increasing where the first derivative is positive, for x < -1 and 0 < x < 1. The function is decreasing elsewhere. It is neither increasing nor decreasing where the first derivative is zero, at x = {-1, 0, 1}, so these points are left out of the intervals for increase or decrease.
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(b) The local extremes are where the first derivative is zero, at x = {-1, 0, 1}. The function values at those points are, respectively, {7, 6, 7}. Hence the outer ones are maxima and the one at (0, 6) is a local minimum. Since the function is an even function that opens downward, the maxima are also global maxima.
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(c) The points of inflection are where the second derivative is zero—where the first derivative has local extremes. The x-values of these points are ±1/√3 = ±(√3)/3, and the y-values are computed to be ...
f(±1/√3) = 6 + 2/3 - 1/9 = 6 5/9
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(d) The graph is concave upward where the first derivative has positive slope (between the points of inflection), and concave downward elsewhere. Of course, it is neither concave upward nor downward at the points of inflection, so those are left out of the intervals.
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Answer:
4, 1, 6, 2, 3, 5
Step-by-step explanation:
When you have so many to do, it is convenient to let a calculator or spreadsheet do the math for you. The attached calculated the sums. Then the magnitudes needed to be calculated from those.
The numbers listed in the answer section are the order in which the resultant vectors appear in the ascending order list. That is, (u+2v-w)/6 appears first in the list.
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Each resultant is found by adding the given vectors with the given coefficients. For example, consider the first one on the list:
... -1/2u +5v = (-1/2)(9, -2) + 5(-1, 7)
... = (-9/2 -5, +1 +35) = (-19/2, 36)
Then the magnitude of it is found by adding the components according to the Pythagorean theorem. For the resultant just computed, the magnitude is ...
... √((-19/2)² +36²) = √(90.25 +1296) = √1386.25 ≈ 37.23
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In the attached, matrix multiplication is used to compute all 6 resultants at once. The 6 resultant vectors are the answer to that multiplication. The magnitudes are computed from these results as above.
I get the magnitudes of the resultants (to 2 decimal places) to be ...
37.23, 3.89, 42.04, 28.50, 32.04, 41.10
