HeLP ME WITH THIS PLEASE!!!!!!

podryga [215]

Critical points is where the derivative (slope) is zero or does not exist. So to do this we have to find the derivative of our function:

$\frac{d}{dx}(x^{2} - 1)^{3}$

So we apply chain rule:

= $3(x^{2} - 1)^{2} * 2x$

Set our first derivative to zero and solve for x:

3(x^2 - 1) * 2x = 0

So we can see that (by plugging in) 0, -1 and 1 makes our solution true

So our critical value is x = 0, x = -1, x = 1

5 0

3 years ago

The change in water level of a lake is modeled by a polynomial function, W(x). Describe how to find the x-intercepts of W(x) and

Agata [3.3K]

First. Finding the x-intercepts of $W(x)$

Let $W(x)$ be the change in water level. So to find the x-intercepts of this function we can use The Rational Zero Test that states:

To find the zeros of the polynomial:

$f(x)=a_{n}x^{n}+a_{n-1}x^{n-1}+...+a_{2}x^{2}+a_{1}x+a_{0}$

We use the Trial-and-Error Method which states that a factor of the constant term:

$a_{0}$

can be a zero of a polynomial (the x-intercepts).

So let's use an example: Suppose you have the following polynomial:

$W(x)=x^{4}-x^{3}-7x^{2}+x+6$

where the constant term is $a_{0}=6$ . The possible zeros are the factors of this term, that is:

$1, -1, 2, -2, 3, -3, 6 \ and \ -6$ .

Thus:

 $W(1)=0 \\ W(-1)=0 \\ W(2)=-12 \\ W(-2)=0 \\ W(3)=0 \\ W(-3)=48 \\ W(6)=840 \\ W(-6)=1260$ 

From the foregoing, we can affirm that $1, -1, -2 \ and \ 3$ are zeros of the polynomial.

Second. Construction a rough graph of $W(x)$

Given that this is a polynomial, then the function is continuous. To graph it we set the roots on the coordinate system. We take the interval:

$[-2,-1]$

and compute $W(c)$ where $c$ is a real number between -2 and -1. If $W(c)>0$ , the curve start rising, if not, the curve start falling. For instance:

$If \ c=-\frac{3}{2} \\ \\ then \ w(-\frac{3}{2})=-2.81$

Therefore the curve start falling and it goes up and down until $x=3$ and from this point it rises without a bound as shown in the figure below

7 0

3 years ago

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-t+5=t-19 Need an answer and a work pls!

Nadusha1986 [10]

Answer:

t=12

Step-by-step explanation:

Step 1: Subtract t from both sides.

−t+5−t=t−19−t

−2t+5=−19

Step 2: Subtract 5 from both sides.

−2t+5−5=−19−5

−2t=−24

Step 3: Divide both sides by -2.

−2t−2=−24−2

8 0

3 years ago

What is the quotient 7^-6/7^2

Serggg [28]

Answer:

7 to the power of negative 8

8 0

2 years ago

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Use spherical coordinates. Find the volume of the solid that lies within the sphere x^2 + y^2 + z^2 = 81, above the xy-plane, an

Natasha_Volkova [10]

Answer:

The volume of the solid is 243 $\sqrt{2} \ \pi$

Step-by-step explanation:

From the information given:

BY applying sphere coordinates:

0 ≤ x² + y² + z² ≤ 81

0 ≤ ρ² ≤ 81

0 ≤ ρ ≤ 9

The intersection that takes place in the sphere and the cone is:

$x^2 +y^2 ( \sqrt{x^2 +y^2 })^2 = 81$

$2(x^2 + y^2) =81$

$x^2 +y^2 = \dfrac{81}{2}$

Thus; the region bounded is: 0 ≤ θ ≤ 2π

This implies that:

$z = \sqrt{x^2+y^2}$

ρcosФ = ρsinФ

tanФ = 1

Ф = π/4

Similarly; in the X-Y plane;

z = 0

ρcosФ = 0

cosФ = 0

Ф = π/2

So here; $\dfrac{\pi}{4} \leq \phi \le \dfrac{\pi}{2}$

Thus, volume: $V = \iiint_E \ d V = \int \limits^{\pi/2}_{\pi/4} \int \limits ^{2\pi}_{0} \int \limits^9_0 \rho ^2 \ sin \phi \ d\rho \ d \theta \ d \phi$

$V = \int \limits^{\pi/2}_{\pi/4} \ sin \phi \ d \phi \int \limits ^{2\pi}_{0} d \theta \int \limits^9_0 \rho ^2 d\rho$

$V = \bigg [-cos \phi \bigg]^{\pi/2}_{\pi/4} \bigg [\theta \bigg]^{2 \pi}_{0} \bigg [\dfrac{\rho^3}{3} \bigg ]^{9}_{0}$

$V = [ -0+ \dfrac{1}{\sqrt{2}}][2 \pi -0] [\dfrac{9^3}{3}- 0 ]$

V = 243 $\sqrt{2} \ \pi$

4 0

3 years ago