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Lyrx [107]
4 years ago
11

Consider the following function. f(x) = (3 − x)e−x (a) Find the intervals of increase or decrease. a. increasing b. decreasing(b

) Find the intervals of concavity. (Enter your answers using interval notation. If an answer does not exist, enter DNE.) a. concave up b. concave down(c) Find the point of inflection. (If an answer does not exist, enter DNE.)(x, y) =
Mathematics
1 answer:
marin [14]4 years ago
5 0

Answer:

a)

f is increasing in the interval (4,+\infty)

f is decreasing in the interval  (-\infty,4)

b)

f is concave up in the interval (-\infty,5)

f is concave down in the interval  (5,+\infty)

c)

x = 5 is the point of inflection.

Step-by-step explanation:

We have

f(x)=(3-x)e^{-x}

(a) Find the intervals of increase or decrease.

To find where the function is increasing, we must find the interval(s) where f'(x)>0

By the rule of the derivative of a product:

f'(x)=-e^{-x}-(3-x)e^{-x}\\\\f'(x)>0\Rightarrow -e^{-x}>(3-x)e^{-x}

since

e^{-x}>0

we divide both sides by

e^{-x}>0

and we get

-1>3-x\Rightarrow x>4

so<em> f is increasing in the interval  </em>

(4,+\infty)

Similarly, we can see <em>f is decreasing in the interval </em>

(-\infty,4)

(b) Find the intervals of concavity.

The function is concave up in the interval(s) where f''>0

f''(x)=2e^{-x}+(3-x)e^{-x}\\\\f''(x)>0\Rightarrow 2e^{-x}+(3-x)e^{-x}>0\Rightarrow 2+3-x>0\Rightarrow x

so <em>f is concave up in the interval </em>

(-\infty,5)

Similarly, we can see <em>f is concave down in the interval </em>

(5,+\infty)

(c) Find the point of inflection.

Since <em>f changes its concavity at x=5</em>, this point is a point of inflection.

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Instructions:Select the correct answer from each drop-down menu. ∆ABC has vertices at A(11, 6), B(5, 6), and C(5, 17). ∆XYZ has
Akimi4 [234]

to compare the triangles, first we will determine the distances of each side

<span>Distance = ((x2-x1)^2+(y2-y1)^2)^0.5
</span>Solving 

<span>∆ABC  A(11, 6), B(5, 6), and C(5, 17)</span>

<span>AB = 6 units   BC = 11 units AC = 12.53 units
</span><span>∆XYZ  X(-10, 5), Y(-12, -2), and Z(-4, 15)
</span><span>XY = 7.14 units   YZ = 18.79 units XZ = 11.66 units</span>

<span>∆MNO  M(-9, -4), N(-3, -4), and O(-3, -15).</span>

<span>MN = 6 units   NO = 11 units MO = 12.53 units
</span><span>∆JKL  J(17, -2), K(12, -2), and L(12, 7).
</span><span>JK = 5 units   KL = 9 units JL = 10.30 units
</span><span>∆PQR  P(12, 3), Q(12, -2), and R(3, -2)
</span><span>PQ = 5 units   QR = 9 units PR = 10.30 units</span> 
Therefore
<span>we have the <span>∆ABC   and the </span><span>∆MNO  </span><span> 
with all three sides equal</span> ---------> are congruent  
</span><span>we have the <span>∆JKL  </span>and the <span>∆PQR 
</span>with all three sides equal ---------> are congruent  </span>

 let's check

 Two plane figures are congruent if and only if one can be obtained from the other by a sequence of rigid motions (that is, by a sequence of reflections, translations, and/or rotations).

 1)     If ∆MNO   ---- by a sequence of reflections and translation --- It can be obtained ------->∆ABC 

<span> then </span>∆MNO<span> ≅</span> <span>∆ABC  </span> 

 a)      Reflexion (x axis)

The coordinate notation for the Reflexion is (x,y)---- >(x,-y)

<span>∆MNO  M(-9, -4), N(-3, -4), and O(-3, -15).</span>

<span>M(-9, -4)----------------->  M1(-9,4)</span>

N(-3, -4)------------------ > N1(-3,4)

O(-3,-15)----------------- > O1(-3,15)

 b)      Reflexion (y axis)

The coordinate notation for the Reflexion is (x,y)---- >(-x,y)

<span>∆M1N1O1  M1(-9, 4), N1(-3, 4), and O1(-3, 15).</span>

<span>M1(-9, -4)----------------->  M2(9,4)</span>

N1(-3, -4)------------------ > N2(3,4)

O1(-3,-15)----------------- > O2(3,15)

 c)   Translation

The coordinate notation for the Translation is (x,y)---- >(x+2,y+2)

<span>∆M2N2O2  M2(9,4), N2(3,4), and O2(3, 15).</span>

<span>M2(9, 4)----------------->  M3(11,6)=A</span>

N2(3,4)------------------ > N3(5,6)=B

O2(3,15)----------------- > O3(5,17)=C

<span>∆ABC  A(11, 6), B(5, 6), and C(5, 17)</span>

 ∆MNO  reflection------- >  ∆M1N1O1  reflection---- > ∆M2N2O2  translation -- --> ∆M3N3O3 

 The ∆M3N3O3=∆ABC 

<span>Therefore ∆MNO ≅ <span>∆ABC   - > </span>check list</span>

 2)     If ∆JKL  -- by a sequence of rotation and translation--- It can be obtained ----->∆PQR 

<span> then </span>∆JKL ≅ <span>∆PQR  </span> 

 d)     Rotation 90 degree anticlockwise

The coordinate notation for the Rotation is (x,y)---- >(-y, x)

<span>∆JKL  J(17, -2), K(12, -2), and L(12, 7).</span>

<span>J(17, -2)----------------->  J1(2,17)</span>

K(12, -2)------------------ > K1(2,12)

L(12,7)----------------- > L1(-7,12)

 e)      translation

The coordinate notation for the translation is (x,y)---- >(x+10,y-14)

<span>∆J1K1L1  J1(2, 17), K1(2, 12), and L1(-7, 12).</span>

<span>J1(2, 17)----------------->  J2(12,3)=P</span>

K1(2, 12)------------------ > K2(12,-2)=Q

L1(-7, 12)----------------- > L2(3,-2)=R

 ∆PQR  P(12, 3), Q(12, -2), and R(3, -2)

 ∆JKL  rotation------- >  ∆J1K1L1  translation -- --> ∆J2K2L2=∆PQR 

<span>Therefore ∆JKL ≅ <span>∆PQR   - > </span><span>check list</span></span>
6 0
4 years ago
Please help the question in on a image down below
enot [183]
The answer for this question is 8 km
8 0
3 years ago
An object is launched at 29.4 meters per
Lerok [7]

Answer:

The reasonable domain for the scenario is option 'a';

a) [0, 7]  

Step-by-step explanation:

For the projectile motion of the object, we are given;

The speed at which the object is launched, v = 29.4 meters per second

The height of the platform from which the object is launched, h = 34.3 meter

The equation for the height of the object as a function of time 'x' is given as follows;

f(x) = -4.9·x² + 29.4·x + 34.3

The domain for the scenario, is given by the possible values of 'x' for the function, which is found as follows;

At the height from which the object is launched, x = 0, and f(x) = 34.3

At the ground level to which the object can drop, f(x) = 0

∴ f(x) = -4.9·x² + 29.4·x + 34.3 = 0

-4.9·x² + 29.4·x + 34.3 = 0

By the quadratic formula, we have;

x = (-29.4 ± √(29.4² - 4 × (-4.9) × 34.3))/(2 × (-4.9)

∴ x = -1, or 7

Given that time is a natural number, we have the reasonable domain for the scenario as the start time when the object is launched, t = 0 to the time the object reaches the ground, t = 7

Therefore, the reasonable domain for the scenario is; 0 ≤ x ≤ 7 or [0, 7].

4 0
3 years ago
Find the value of X and y<br>☺ -3/2y = -13<br>6X2Y = 4<br>by elimination​
alina1380 [7]
X equals 13/3 and Y equals 1/13 sorry if I am wrong
4 0
3 years ago
What is the slope of the line through (-1, 2) and (-3, -2)
Juli2301 [7.4K]

2

Step-by-step explanation:

-2 -2/ -3+1 = -4/-2 = 2

you just have to use the slope formula

6 0
3 years ago
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