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frozen [14]
3 years ago
10

Select wether the equation has a solution or not

Mathematics
2 answers:
Leona [35]3 years ago
7 0

Answer:

No roots

Step-by-step explanation:

Umnica [9.8K]3 years ago
3 0

Answer:

NO ROOTS

Step-by-step explanation:

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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}.

7 0
3 years ago
Find the nth term:
san4es73 [151]

Answer:

n+3

Step-by-step explanation:

the relationship between those numbers are consistently adding 3 thus n +3

7 0
3 years ago
Read 2 more answers
Find the volume of each cylinder. Use 3.14 for pi. Round your answer to the nearest tenth
otez555 [7]

Answer:

V ≈471.24 mm^3

Step-by-step explanation:

The formula for cylinder volume is πr^2 x h, so ((π x 25) x h). That's just 25π x 6. That is about 471.238898, which rounded is almost 471.24. Or, in terms of π, you could leave your answer as 150π mm^3

8 0
3 years ago
Please help me im stuck on this question​
STALIN [3.7K]

Answer:

78.0 kilo

Step-by-step explanation:

85.8- 93.6= 7.8

101.4-93.6= 7.8

85.8-7.8 = 78.0 kilo

6 0
3 years ago
Can someone please help me on this question?​
yan [13]

Answer: q³⁰

Explanation:

First just solve the first part using the exponent rules

p²q⁵ becomes 1/p-⁸q-²⁰ then we flip the fraction so the exponents become positive. Now we have p⁸q²⁰.

Before multiplying the other equation, we must simplify. p-⁴q⁵ becomes 1/p⁴q-⁵ and since it's the exponents being raised to a power we simply multiply the inner exponents times the outer exponent which yields 1/p⁸q-¹⁰. We must make q-¹⁰ positive so we will then bring it to the numerator of the fraction which gives us: q¹⁰/p⁸.

Multiply q¹⁰/p⁸ * p⁸q²⁰/1 = p⁸q³⁰/p⁸ divide the p exponents by each other which yields 0 since when u divide exponents you just subtract them so 8 - 8 = 0. Your answer is now q³⁰/1 or just q³⁰

4 0
3 years ago
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