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Strike441 [17]
3 years ago
11

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.

Mathematics
1 answer:
exis [7]3 years ago
7 0

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.

___

(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.

__

(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.

__

(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

__

(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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Hi there!

<u><em>FACTS</em></u><em> :</em>

<em>To see if multiple ratios are proportional, you can write them as fractions, reduce them, and compare them. If the reduced fractions are all the same, then you have proportionnal ratios. You can also write them as fractions and divide the numerator (top number) by its denominator (bottom number), and compare the decimal numbers the same way you would compare the fractions (I personnaly find this method easier because you don't need to simplify the fractions).</em>

<u>STEPS TO ANSWER:</u>

x = 1 ; y = 90 → \frac{1}{90} = 1 ÷ 90 = <u>0.01111111...</u>


x = 2 ; y = 150 → \frac{2}{150} = 2 ÷ 150 = <u>0.0133333...</u>


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x = 4 ; y = 270 → \frac{4}{270} = 4 ÷ 270 = <u>0.0148148...</u>


x = 5 ; y = 330 → \frac{5}{330} = 5 ÷ 330 = <u>0.01515152</u>


<em>** You didn't really need to calculate them all because even the first two decimal numbers weren't equivalent, but I wanted to show you the whole process so I calculated them all.</em>


⇒ If you compare all the decimals you got, you can easily see that they are not the same, which means that the ratios between the values of "x" and the values of "y" are not proportional. Therefore, George's reasoning wasn't good.


There you go! I really hope this helped, if there's anything just let me know! :)

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