CL ≈ AE and HG≈ZR and ∠L≈∠E as both triangles are isosceles so CL=HL and AE=EZ
so HL≈AE≈ZR≈EZ so according to side angle side ΔCLH andΔ AZE both are congruent so side CH≈AZ
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.
You multiply like terms first, which means -2/3x*-x. It becomes 2/3x*3y. The coefficient of x is 2/3. The coefficient of y is 3. You can’t combine them because they are not like terms with the variables.
He can either measure the third side length, apply the Pythagorean theorem to find the height of the triangle, and then calculate the area, or he can find the measure of the included angle between the known side lengths and use trigonometry to express the height of the triangle and then determine the
area of the triangle
