Question

multiple choice 2 points
what is the value of w in the equation \\( \frac { 3 } { 4 } w + 8 = w - 7 \\)?
-0.2
60
2.4
-13.846
multiple choice 2 points
let \\( f ( x ) = ( x ^ { 2 } - 1 ) ^ { 4 } \\). over what interval is the function decreasing?
\\( ( 1 , \infty ) \\)
\\( ( - \infty , - 1 ) \\) and \\( ( 0,1 ) \\)
\\( ( - 1,1 ) \\)
\\( ( - 1,0 ) \\) and \\( ( 1 , \infty ) \\)

Explanation:

Question 28

Step1: Solve the equation

Given \(\frac{3}{4}w + 8=w - 7\).
First, move the terms with \(w\) to one side:
\(8 + 7=w-\frac{3}{4}w\).

Step2: Simplify the right - hand side

\(15=\frac{4w-3w}{4}\), since \(w=\frac{4w}{4}\).
So \(15=\frac{w}{4}\).

Step3: Solve for \(w\)

Multiply both sides by \(4\): \(w = 60\).

Step1: Find the derivative of \(y=(x^{2}-1)^{4}\)

Let \(u = x^{2}-1\), then \(y = u^{4}\).
Using the chain rule \(\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}\).
\(\frac{dy}{du}=4u^{3}\) and \(\frac{du}{dx}=2x\), so \(\frac{dy}{dx}=4(x^{2}-1)^{3}\cdot2x=8x(x^{2}-1)^{3}\).

Step2: Find where \(\frac{dy}{dx}<0\)

Set \(\frac{dy}{dx}=8x(x - 1)^{3}(x + 1)^{3}<0\).
We can use a sign - chart.
Test intervals:

• For \(x\in(-\infty,- 1)\), let \(x=-2\). Then \(\frac{dy}{dx}=8\times(-2)\times((-2)^{2}-1)^{3}=8\times(-2)\times(3)^{3}<0\).
• For \(x\in(-1,0)\), let \(x =-\frac{1}{2}\). Then \(\frac{dy}{dx}=8\times(-\frac{1}{2})\times((-\frac{1}{2})^{2}-1)^{3}=8\times(-\frac{1}{2})\times(-\frac{3}{4})^{3}>0\).
• For \(x\in(0,1)\), let \(x=\frac{1}{2}\). Then \(\frac{dy}{dx}=8\times\frac{1}{2}\times((\frac{1}{2})^{2}-1)^{3}=8\times\frac{1}{2}\times(-\frac{3}{4})^{3}<0\).
• For \(x\in(1,\infty)\), let \(x = 2\). Then \(\frac{dy}{dx}=8\times2\times((2)^{2}-1)^{3}=8\times2\times(3)^{3}>0\).

Answer:

\(60\)

Question 29