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lutik1710 [3]
2 years ago
9

Consider the function y=3x^5-25x^3+60x+1. Use the first or second derivative test to test the critical points. How many relative

maxima did you find? a)0 b)1 c)2 d)3 e)4
Mathematics
1 answer:
Vilka [71]2 years ago
6 0
1) First derivative

dy / dx = 15x^4 - 75x^2 + 60

relative maxima => dy / dx = 0

=> 15x^4 - 75x^2 + 60 = 0

=> x^4 - 5x^2 + 4 = 0

Solve by factoring: (x^2 - 4) (x^2 - 1) = 0

=> x^2 = 4 => x = 2 and x = -2

x^2 = 1 => x = 1 and x = -1

So, there are four candidates -4, -1, 1 and 4.

To conclude whether they are relative maxima you must take the second derivative. Only those whose second derivative is negative are relative maxima.

2) Second derivative:

d^2 y / dx^2 = 60x^3 - 150x

Factor: 30x(2x^2 - 5)

The second derivative test tells that the second derivative in a relative maxima is less than zero.

=> 30x (2x^2 - 5) < 0

=> x < 0 and x^2 > 5/2

 => x < 0 and x < √(5/2) or x > √(5/2)

 => x = - 2 is a local maxima


Also, x > 0 and x^2 < 5/2 is a solution

=> x = 1 is other local maxima.

Those are the only two local maxima.

The answer is the option c) 2 local maxima.

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This is finding exact values of sin theta/2 and tan theta/2. I’m really confused and now don’t have a clue on how to do this, pl
Lostsunrise [7]

First,

tan(<em>θ</em>) = sin(<em>θ</em>) / cos(<em>θ</em>)

and given that 90° < <em>θ </em>< 180°, meaning <em>θ</em> lies in the second quadrant, we know that cos(<em>θ</em>) < 0. (We also then know the sign of sin(<em>θ</em>), but that won't be important.)

Dividing each part of the inequality by 2 tells us that 45° < <em>θ</em>/2 < 90°, so the half-angle falls in the first quadrant, which means both cos(<em>θ</em>/2) > 0 and sin(<em>θ</em>/2) > 0.

Now recall the half-angle identities,

cos²(<em>θ</em>/2) = (1 + cos(<em>θ</em>)) / 2

sin²(<em>θ</em>/2) = (1 - cos(<em>θ</em>)) / 2

and taking the positive square roots, we have

cos(<em>θ</em>/2) = √[(1 + cos(<em>θ</em>)) / 2]

sin(<em>θ</em>/2) = √[(1 - cos(<em>θ</em>)) / 2]

Then

tan(<em>θ</em>/2) = sin(<em>θ</em>/2) / cos(<em>θ</em>/2) = √[(1 - cos(<em>θ</em>)) / (1 + cos(<em>θ</em>))]

Notice how we don't need sin(<em>θ</em>) ?

Now, recall the Pythagorean identity:

cos²(<em>θ</em>) + sin²(<em>θ</em>) = 1

Dividing both sides by cos²(<em>θ</em>) gives

1 + tan²(<em>θ</em>) = 1/cos²(<em>θ</em>)

We know cos(<em>θ</em>) is negative, so solve for cos²(<em>θ</em>) and take the negative square root.

cos²(<em>θ</em>) = 1/(1 + tan²(<em>θ</em>))

cos(<em>θ</em>) = - 1/√[1 + tan²(<em>θ</em>)]

Plug in tan(<em>θ</em>) = - 12/5 and solve for cos(<em>θ</em>) :

cos(<em>θ</em>) = - 1/√[1 + (-12/5)²] = - 5/13

Finally, solve for sin(<em>θ</em>/2) and tan(<em>θ</em>/2) :

sin(<em>θ</em>/2) = √[(1 - (- 5/13)) / 2] = 3/√(13)

tan(<em>θ</em>/2) = √[(1 - (- 5/13)) / (1 + (- 5/13))] = 3/2

3 0
2 years ago
Can someone plz help me
PtichkaEL [24]

Answer:

Your answer is exactly 120 feet

3 0
2 years ago
Read 2 more answers
Need answers plz help
stiv31 [10]
You can use the factors of the volumes 24, 27 and 48:
For example:
8 by 3 by 1 is a total volume of 24
or if you know that 4 times 2 is 8:
4 by 2 by 2
and so on
3 0
3 years ago
(Ross 5.15) If X is a normal random variable with parameters µ " 10 and σ 2 " 36, compute (a) PpX ą 5q (b) Pp4 ă X ă 16q (c) PpX
pochemuha

Answer:

(a) 0.7967

(b) 0.6826

(c) 0.3707

(d) 0.9525

(e) 0.1587

Step-by-step explanation:

The random variable <em>X</em> follows a Normal distribution with mean <em>μ</em> = 10 and  variance <em>σ</em>² = 36.

(a)

Compute the value of P (X > 5) as follows:

P(X>5)=P(\frac{x-\mu}{\sigma}>\frac{5-10}{\sqrt{36}})\\=P(Z>-0.833)\\=P(Z

Thus, the value of P (X > 5) is 0.7967.

(b)

Compute the value of P (4 < X < 16) as follows:

P(4

Thus, the value of P (4 < X < 16) is 0.6826.

(c)

Compute the value of P (X < 8) as follows:

P(X

Thus, the value of P (X < 8) is 0.3707.

(d)

Compute the value of P (X < 20) as follows:

P(X

Thus, the value of P (X < 20) is 0.9525.

(e)

Compute the value of P (X > 16) as follows:

P(X>16)=P(\frac{x-\mu}{\sigma}>\frac{16-10}{\sqrt{36}})\\=P(Z>1)\\=1-P(Z

Thus, the value of P (X > 16) is 0.1587.

**Use a <em>z</em>-table for the probabilities.

8 0
3 years ago
Si f(x) es igual a raiz de x al cuadrado menos 16 determinar f(5),f(4),f(6),f(3)
Nadusha1986 [10]

Aplicando la función, los valores numéricos son :

f(5) = -11

f(4) = -12

f(6) = -10

f(3) = -13

La función es dada por:

f(x) = (\sqrt{x})^2 - 16  = x - 16

Para los valores numéricos, reemplazamos x, luego:

f(5) = 5 - 16 = -11

f(4) = 4 - 16 = -12

f(6) = 6 - 16 = -10

f(3) = 3 - 16 = -13

Un problema similar es dado en brainly.com/question/7037337

4 0
2 years ago
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