relative maxima and minima,唔識!

lam

2010-09-15 5:44 am

1. Given the function f(x)=ax^3+2ax^2+ax+1, complete the following.

Find the relative maximum value of f(x) when a<0.

2. Given the function f(x)=ax^3+3ax^2-9ax+1, complete the following.

Find the relative maximum value of f(x) when a>0.

Find the relative maximum value of f(x) when a<0.

3. Find the relative minimum value of the function f(x)=ax^3-3ax+4,
where a「不等於」0. (Consider the cases where a>0 and a<0)

4. Find the relative maximum value of the function f(x)=ax^3-2ax^2-4ax+8
(where a「不等於」0)

5. Given that the function f(x)=ax^3+3ax^2-9ax(where a「不等於」0) has a
relative maximum value of 15, complete the following.

Find the relative minimum value when a<0.

6. Given that the function f(x)=ax^3+9ax^2+24ax-4 (where a「不等於」0) has a
relative maximum value of -20, complete the following.

回答 (1)

用戶10012

2010-09-15 4:04 pm

✔ 最佳答案

All 6 questions are quite similar in nature, so I just work on Q2, you can follow the steps to do the other questions.
2. f = ax^3 + 3ax^2 - 9ax + 1
Step 1 : Find df/dx.
df/dx = 3ax^2 + 6ax - 9a
Step 2: Put df/dx = 0 in order to find the value of x of the relative maximum or minimum.
df/dx = 3ax^2 + 6ax - 9a = 0
x^2 + 2x - 3 = 0
(x +3)(x - 1) = 0
so x = 1 or - 3.
Step 3 : Determine which one is max. ( or min.) by the testing method f" < 0 is a max. ( f"> 0 is a min.)
f" = d(3ax^2 + 6ax - 9a)/dx = 6ax + 6a = 6a( x + 1).
When a > 0 and f(x) has to be a max. (f" < 0). This happens when x = - 3.
so f(x) max.= a(-3)^3 + 3a(-3)^2 - 9a(-3) + 1 = - 27a + 27a + 27a + 1 = 27a + 1.
When a < 0 and f(x) has to be a max. (f"< 0). This happens when x = 1.
so f(x) max. = a(1)^3 + 3a(1)^2 - 9a(1) + 1 = a + 3a - 9a + 1 = 1 - 5a.

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