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babunello [35]
3 years ago
13

F(x) = x4 - 32x2 + 7

Mathematics
1 answer:
xenn [34]3 years ago
4 0
F is increasing if f '(x) = 4x^3<span> - 64x >0, so it is the same of  x(x^2 -16)>0, implies x>0 or (x-4)(x+4)>0, so x> + or -4
the answer is </span><span>(-4 , 0) ∪ (4 , infinity)

</span>f is decreasing if f '(x) = 4x^3 - 64x <0, so it is the same of  x(x^2 -16)<0, implies x<0 or (x-4)(x+4)<0, so x< + or -4
the answer is (- infinity, -4) ∪ (0 , 4)

<span>the local minimum and maximum
</span>f '(x) =0, impolies x=+ or -4, or x=0
or f'(o)=0,  and  f'(- 4)=f'(4)= 0, 
M(-4, 0) or M(4, 0) or M(0,0)

<span>inflection points can be found by solving f '' (x)=12x^2 - 64 =0
</span>x=+ or - 4sqrt(3) / 3
so the inflection point is  and M(- 4sqrt(3) / 3, f'' ( -4sqrt(3))  (smaller x value), and M(4sqrt(3) / 3, f'' (4sqrt(3)) (larger x value)

f is concave up if f ''>0
it means 12x^2 - 64>0, so the interval is (- infinity, -4sqrt(3) U (4sqrt(3), infinity)

f is concave down if f ''<0
it means 12x^2 - 64<0 so the interval is (-4sqrt(3)) U (4sqrt(3))

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I need help on this.​
SSSSS [86.1K]

Answer:

plot the first point on (0,-4)

from there, rise 2, then run 3

plot another point on (3,-2)

plot another on (6,0), and so on

<em><u>Plz mark as brainliest if correct! Have a nice day!!! </u></em>

<em><u>-Lil G</u></em>

8 0
3 years ago
rachel was cutting out some fabric for a friend. she cut a piece that was 5 cm wide and had an area of 20(20)cm. how long was th
Pavlova-9 [17]
Area of a rectangle is L * W

L = x
W = 5

5x = 20 (Divide by 5 on both sides)
x = 4 

L=4
W= 5

The length is 4cm 

Hope this helps :)
3 0
3 years ago
Let f be defined by the function f(x) = 1/(x^2+9)
riadik2000 [5.3K]

(a)

\displaystyle\int_3^\infty \frac{\mathrm dx}{x^2+9}=\lim_{b\to\infty}\int_{x=3}^{x=b}\frac{\mathrm dx}{x^2+9}

Substitute <em>x</em> = 3 tan(<em>t</em> ) and d<em>x</em> = 3 sec²(<em>t </em>) d<em>t</em> :

\displaystyle\lim_{b\to\infty}\int_{t=\arctan(1)}^{t=\arctan\left(\frac b3\right)}\frac{3\sec^2(t)}{(3\tan(t))^2+9}\,\mathrm dt=\frac13\lim_{b\to\infty}\int_{t=\arctan(1)}^{t=\arctan\left(\frac b3\right)}\mathrm dt

=\displaystyle \frac13 \lim_{b\to\infty}\left(\arctan\left(\frac b3\right)-\arctan(1)\right)=\boxed{\dfrac\pi{12}}

(b) The series

\displaystyle \sum_{n=3}^\infty \frac1{n^2+9}

converges by comparison to the convergent <em>p</em>-series,

\displaystyle\sum_{n=3}^\infty\frac1{n^2}

(c) The series

\displaystyle \sum_{n=1}^\infty \frac{(-1)^n (n^2+9)}{e^n}

converges absolutely, since

\displaystyle \sum_{n=1}^\infty \left|\frac{(-1)^n (n^2+9)}{e^n}\right|=\sum_{n=1}^\infty \frac{n^2+9}{e^n} < \sum_{n=1}^\infty \frac{n^2}{e^n} < \sum_{n=1}^\infty \frac1{e^n}=\frac1{e-1}

That is, ∑ (-1)ⁿ (<em>n</em> ² + 9)/<em>e</em>ⁿ converges absolutely because ∑ |(-1)ⁿ (<em>n</em> ² + 9)/<em>e</em>ⁿ| = ∑ (<em>n</em> ² + 9)/<em>e</em>ⁿ in turn converges by comparison to a geometric series.

5 0
3 years ago
Write an example of a trinomial that cannot be factored.
OverLord2011 [107]

Answer:

x^{2} + x + 1

Step-by-step explanation:

a trinomial that cannot be factored has no real roots.

here is one:

x^{2} + x + 1

After graphing this trinomial has no intersections points with the x-axis wich means it has no roots

5 0
3 years ago
1.1 solve for x, where 0°&lt; x 90°. write your answer to one decimal place. 1.1.1 tanx=sin38° 1.1.2cosec( x+10°)=1.345​
snow_lady [41]

The value of x in tan(x)=sin38° is 31.6 and the value of x in cosec(x+10°)=1.345​ is 38.0

<h3>How to solve the trigonometry ratios?</h3>

The equations are given as:

tan(x)=sin38°

cosec( x+10°)=1.345​

In tan(x)=sin38°, we have:

tan(x)=0.6157

Take the arc tan of both sides

x = 31.6

Also, we have:

cosec(x+10°)=1.345​

Take the inverse of both sides

sin(x+10°) = 0.7434

Take the arc sin of both sides

x+10 = 48.0

Subtract 10 from both sides

x = 38.0

Hence, the value of x in tan(x)=sin38° is 31.6 and the value of x in cosec(x+10°)=1.345​ is 38.0

Read more about trigonometry ratios at:

brainly.com/question/11967894

#SPJ1

5 0
2 years ago
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