49 multiple choice 2 points\nwhat is the value of x in the solution of the system of equations ( 3x + 2y =…

Answer:

49. C. 2
50. B. ((-1,0)) and ((1,\infty))

Explanation:

49. Solving the system of equations (3x + 2y=12) and (5x - 2y = 4)

Step1: Add the two equations

Adding (3x + 2y=12) and (5x - 2y = 4) to eliminate (y). ((3x + 2y)+(5x - 2y)=12 + 4) (3x+5x+2y - 2y=16) (8x=16)

Step2: Solve for (x)

Divide both sides of (8x = 16) by (8). (x=\frac{16}{8}=2)

50. Finding where (f(x)=(x^{2}-1)^{4}) is increasing

Step1: Find the derivative using the chain - rule

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

Step2: Find the critical points

Set (f^{\prime}(x)=0). Then (8x(x - 1)^{3}(x + 1)^{3}=0), so (x=-1,0,1)

Step3: Use the test - intervals

• For (x\in(-\infty,-1)), let (x=-2). Then (f^{\prime}(-2)=8\times(-2)\times((-2)-1)^{3}\times((-2)+1)^{3}=8\times(-2)\times(-27)\times(-1)<0)
• For (x\in(-1,0)), let (x =-\frac{1}{2}). Then (f^{\prime}(-\frac{1}{2})=8\times(-\frac{1}{2})\times((-\frac{1}{2})-1)^{3}\times((-\frac{1}{2})+1)^{3}=8\times(-\frac{1}{2})\times(-\frac{27}{8})\times(\frac{1}{8})>0)
• For (x\in(0,1)), let (x=\frac{1}{2}). Then (f^{\prime}(\frac{1}{2})=8\times\frac{1}{2}\times((\frac{1}{2})-1)^{3}\times((\frac{1}{2})+1)^{3}=8\times\frac{1}{2}\times(-\frac{1}{8})\times(\frac{27}{8})<0)
• For (x\in(1,\infty)), let (x = 2). Then (f^{\prime}(2)=8\times2\times(2 - 1)^{3}\times(2 + 1)^{3}=8\times2\times1\times27>0)

The function (f(x)) is increasing when (f^{\prime}(x)>0), which is on the intervals ((-1,0)) and ((1,\infty))