find the absolute maximum absolute minimum of the function F(x)=(x^2-12x+32)^(1/3) on the interval [3,10]
1 answer:
9514 1404 393
Answer:
- minimum: -∛4 ≈ -1.58740
- maximum: ∛12 ≈ 2.28943
Step-by-step explanation:
The expression under the radical can be written in vertex form as ...
x^2 -12x +32 = (x -6)^2 -4
The function will be symmetrical about x=6. The minimum is at x=6 where we have ...
F(6) = ∛(-4)
The maximum is at the end of the interval farthest from the vertex, so is ...
F(10) = ∛((10 -6)² -4) = ∛12
The minimum is -∛4 ≈ -1.587; the maximum is ∛12 ≈ 2.289.
You might be interested in
Answer:
- 22
Step-by-step explanation:
A fall of degree = 0.5 degrees is represented as
- 0. 5 degrees
The temperature at 2 : 00 is therefore
- 21.5 - 0.5 = - 22 degrees
Answer:
(-4,2) belongs in this direct variation.
Step-by-step explanation:
Let the direct variation relationship is expressed by the equation y = mx ........ (1), where x and y are in direct variation and k is the variation constant.
Now, the point (4,-2) is included in the direct variation relationship, then from equation (1) we get, -2 = 4m
⇒
Therefore, the equation (1) becomes
⇒ x + 2y = 0 ......... (2)
Now, from the given four options only the point (-4,2) satisfies the relation (2).
Hence, (-4,2) belongs in this direct variation. (Answer)