Answer:
(x,y) ⇒ (x - 3) (y - 4)
C(5, -1)
Step-by-step explanation:
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.
![\begin{cases} 4x+3y=-8\\\\ -8x-6y=16 \end{cases}~\hspace{10em} \begin{array}{|c|ll} \cline{1-1} slope-intercept~form\\ \cline{1-1} \\ y=\underset{y-intercept}{\stackrel{slope\qquad }{\stackrel{\downarrow }{m}x+\underset{\uparrow }{b}}} \\\\ \cline{1-1} \end{array} \\\\[-0.35em] ~\dotfill](https://tex.z-dn.net/?f=%5Cbegin%7Bcases%7D%204x%2B3y%3D-8%5C%5C%5C%5C%20-8x-6y%3D16%20%5Cend%7Bcases%7D~%5Chspace%7B10em%7D%20%5Cbegin%7Barray%7D%7B%7Cc%7Cll%7D%20%5Ccline%7B1-1%7D%20slope-intercept~form%5C%5C%20%5Ccline%7B1-1%7D%20%5C%5C%20y%3D%5Cunderset%7By-intercept%7D%7B%5Cstackrel%7Bslope%5Cqquad%20%7D%7B%5Cstackrel%7B%5Cdownarrow%20%7D%7Bm%7Dx%2B%5Cunderset%7B%5Cuparrow%20%7D%7Bb%7D%7D%7D%20%5C%5C%5C%5C%20%5Ccline%7B1-1%7D%20%5Cend%7Barray%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill)
![4x+3y=-8\implies 3y=-4x-8\implies y=\cfrac{-4x-8}{3}\implies y=\stackrel{\stackrel{m}{\downarrow }}{-\cfrac{4}{3}} x-\cfrac{8}{3} \\\\[-0.35em] ~\dotfill\\\\ -8x-6y=16\implies -6y=8x+16\implies y=\cfrac{8x+16}{-6} \\\\\\ y=\cfrac{8}{-6}x+\cfrac{16}{-6}\implies y=\stackrel{\stackrel{m}{\downarrow }}{-\cfrac{4}{3}} x-\cfrac{8}{3}](https://tex.z-dn.net/?f=4x%2B3y%3D-8%5Cimplies%203y%3D-4x-8%5Cimplies%20y%3D%5Ccfrac%7B-4x-8%7D%7B3%7D%5Cimplies%20y%3D%5Cstackrel%7B%5Cstackrel%7Bm%7D%7B%5Cdownarrow%20%7D%7D%7B-%5Ccfrac%7B4%7D%7B3%7D%7D%20x-%5Ccfrac%7B8%7D%7B3%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20-8x-6y%3D16%5Cimplies%20-6y%3D8x%2B16%5Cimplies%20y%3D%5Ccfrac%7B8x%2B16%7D%7B-6%7D%20%5C%5C%5C%5C%5C%5C%20y%3D%5Ccfrac%7B8%7D%7B-6%7Dx%2B%5Ccfrac%7B16%7D%7B-6%7D%5Cimplies%20y%3D%5Cstackrel%7B%5Cstackrel%7Bm%7D%7B%5Cdownarrow%20%7D%7D%7B-%5Ccfrac%7B4%7D%7B3%7D%7D%20x-%5Ccfrac%7B8%7D%7B3%7D)
one simple way to tell if both equations do ever meet or have a solution is by checking their slope, notice in this case the slopes are the same for both, meaning the lines are parallel lines, however, notice both equations are really the same, namely the 2nd equation is really the 1st one in disguise.
since both equations are equal, their graph will be of one line pancaked on top of the other, and the solutions is where they meet, hell, they meet everywhere since one is on top of the other, so infinitely many solutions.
We have the following equation:
<span> h(t)=-4.92t^2+17.69t+575
</span> For the domain we have:
<span> </span>We match zero:
-4.92t ^ 2 + 17.69t + 575 = 0
We look for the roots:
t1 = -9.16
t2 = 12.76
We are left with the positive root, so the domain is:
[0, 12.76]
For the range we have:
We derive the function:
h '(t) = - 9.84t + 17.69
We equal zero and clear t:
-9.84t + 17.69 = 0
t = 17.69 / 9.84
t = 1.80
We evaluate the time in which it reaches the maximum height in the function:
h (1.80) = - 4.92 * (1.80) ^ 2 + 17.69 * (1.80) +575
h (1.80) = 590.90
Therefore, the range is given by:
[0, 590.9]
Answer:
the domain and range are:
domain: [0, 12.76] range: [0, 590.9]
