Answer:
b) g(x) is stretched horizontally and reflected over the x-axis.
Step-by-step explanation:
The given parent function is
The transformed function is
We want to see how the transformed graph compares with the parent graph.
The negation inside means all x-coordinates we're negated.
This means, there is a reflection in the y-axis.
The factor of 2 within the function means a horizontal stretch by a factor of 1/2.
The correct answer is B.
Part A. You have the correct first and second derivative.
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Part B. You'll need to be more specific. What I would do is show how the quantity (-2x+1)^4 is always nonnegative. This is because x^4 = (x^2)^2 is always nonnegative. So (-2x+1)^4 >= 0. The coefficient -10a is either positive or negative depending on the value of 'a'. If a > 0, then -10a is negative. Making h ' (x) negative. So in this case, h(x) is monotonically decreasing always. On the flip side, if a < 0, then h ' (x) is monotonically increasing as h ' (x) is positive.
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Part C. What this is saying is basically "if we change 'a' and/or 'b', then the extrema will NOT change". So is that the case? Let's find out
To find the relative extrema, aka local extrema, we plug in h ' (x) = 0
h ' (x) = -10a(-2x+1)^4
0 = -10a(-2x+1)^4
so either
-10a = 0 or (-2x+1)^4 = 0
The first part is all we care about. Solving for 'a' gets us a = 0.
But there's a problem. It's clearly stated that 'a' is nonzero. So in any other case, the value of 'a' doesn't lead to altering the path in terms of finding the extrema. We'll focus on solving (-2x+1)^4 = 0 for x. Also, the parameter b is nowhere to be found in h ' (x) so that's out as well.
Answer:
D
Step-by-step explanation:
1. The domain is all x values for the function.
2. Since the line starts at the x value of 3, it is included and since it goes to the right, it also includes all the values greater than 3.
3. Therefore, x>_3
Square root 50 does not belong....all of the other numbers are rational numbers. square root 50 is irrational.
