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Damm [24]
3 years ago
14

Which of the following statements is not true?

Mathematics
1 answer:
I am Lyosha [343]3 years ago
8 0
I think B isn’t true
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a 40 ounce bag

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Consider the equation below. (If you need to use -[infinity] or [infinity], enter -INFINITY or INFINITY.)f(x) = 2x3 + 3x2 − 180x
soldier1979 [14.2K]

Answer:

(a) The function is increasing \left(-\infty, -6\right) \cup \left(5, \infty\right) and decreasing \left(-6, 5\right)

(b) The local minimum is x = 5 and the maximum is x = -6

(c) The inflection point is x = -\frac{1}{2}

(d) The function is concave upward on \left(- \frac{1}{2}, \infty\right) and concave downward on \left(-\infty, - \frac{1}{2}\right)

Step-by-step explanation:

(a) To find the intervals where f(x) = 2x^3 + 3x^2 -180x is increasing or decreasing you must:

1. Differentiate the function

\frac{d}{dx}f(x) =\frac{d}{dx}(2x^3 + 3x^2 -180x) \\\\\mathrm{Apply\:the\:Sum/Difference\:Rule}:\quad \left(f\pm g\right)'=f\:'\pm g'\\\\f'(x)=\frac{d}{dx}\left(2x^3\right)+\frac{d}{dx}\left(3x^2\right)-\frac{d}{dx}\left(180x\right)\\\\f'(x) =6x^2+6x-180

2. Now we want to find the intervals where f'(x) is positive or negative. This is done using critical points, which are the points where f'(x) is either 0 or undefined.

f'(x) =6x^2+6x-180 =0\\\\6x^2+6x-180 = 6\left(x-5\right)\left(x+6\right)=0\\\\x=5,\:x=-6

These points divide the number line into three intervals:

(-\infty,-6), (-6,5), and (5, \infty)

Evaluate f'(x) at each interval to see if it's positive or negative on that interval.

\left\begin{array}{cccc}Interval&x-value&f'(x)&Verdict\\(-\infty,-6)&-7&72&Increasing\\(-6,5)&0&-180&Decreasing\\(5, \infty)&6&72&Increasing\end{array}\right

Therefore f(x) is increasing \left(-\infty, -6\right) \cup \left(5, \infty\right) and decreasing \left(-6, 5\right)

(b) Now that we know the intervals where f(x) increases or decreases, we can find its extremum points. An extremum point would be a point where f(x) is defined and f'(x) changes signs.

We know that:

  • f(x) increases before x = -6, decreases after it, and is defined at x = -6. So f(x) has a relative maximum point at x = -6.
  • f(x) decreases before x = 5, increases after it, and is defined at x = 5. So f(x) has a relative minimum point at x = 5.

(c)-(d) An Inflection Point is where a curve changes from Concave upward to Concave downward (or vice versa).

Concave upward is when the slope increases and concave downward is when the slope decreases.

To find the inflection points of f(x), we need to use the f''(x)

f''(x)=\frac{d}{dx}\left(6x^2+6x-180\right)\\\\\mathrm{Apply\:the\:Sum/Difference\:Rule}:\quad \left(f\pm g\right)'=f\:'\pm g'\\\\f''(x)=\frac{d}{dx}\left(6x^2\right)+\frac{d}{dx}\left(6x\right)-\frac{d}{dx}\left(180\right)\\\\f''(x) =12x+6

We set f''(x) = 0

f''(x) =12x+6 =0\\\\x=-\frac{1}{2}

Analyzing concavity, we get

\left\begin{array}{cccc}Interval&x-value&f''(x)\\(-\infty,-1/2)&-2&-18\\(-1/2,\infty)&0&6\\\end{array}\right

The function is concave upward on (-1/2,\infty) because the f''(x) > 0 and concave downward on (-\infty,-1/2) because the f''(x) < 0.

f(x) is concave down before x = -\frac{1}{2}, concave up after it. So f(x) has an inflection point at x = -\frac{1}{2}.

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i have been struggling for this work that has been assigned to me at the first day of school after a long time.
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Answer:

Fill the gaps in this sequence:

25 36 <u>49</u> 64 81 <u>100</u>

This sequence is the numbers 5, 6, 7, 8, 9, and 10 squared.

Work out the cube root of 343, then square it.

<u>49</u>

Since 7 cubed equals 343, then 7 squared is 49.

When you subtract one square number from another the answer is 7.

What are the two square numbers?

<u>16</u> and <u>9</u>

These are the only squares that fit the criteria.

Write down a number you can square to give an answer bigger than 200 but smaller than 300.

<u>15, 16, 17</u>

All of those answers work. 15² is 225, 16² is 256, and 17² is 289. You can choose any of them to enter in and it should work.

7 0
3 years ago
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