Question

Help please :)

Let f(x)=54−3x−x^2. Find the open intervals on which f is increasing (decreasing). Then determine the x-coordinates of all relative maxima (minima).

1. f is increasing on the intervals
2. f is decreasing on the intervals
3. The relative maxima of f occur at x =
4. The relative minima of f occur at x =

Notes: In the first two, your answer should either be a single interval, such as (0,1), a comma separated list of intervals, such as (-inf, 2), (3,4), or the word "none"

Answer 1

since it is a quadratic with negative leading coefficient, it opens down
therefore it is increasing from \(-\infty\) to the first coordinate of the vertex \(-\frac{b}{2a}\) and decreasing from there on

Answer 2

@dan815

Answer 3

ok so

1. would be from -infinity to the result from -b/2a?

Answer 4

yes

Answer 5

"a" would be 1 and "b" would be -3?

Answer 6

no

Answer 7

\[f(x)=54−3x−x^2.\]
\[f(x)=-x^2-3x+54\\
f(x)=ax^2+bx+c\]

Answer 8

ohh so -1 and -3?

Answer 9

yes

Answer 10

@satellite73 I typed for the first one (-inf,-.1666) and im getting it wrong

Answer 11

@aum help!

Answer 12

The function increases from x = -infinity to x = -b/2a = -(-3) / (2*-1) = 3/(-2) = -1.5
The interval is: (-infinity, -1.5)

Answer 13

@aum so what would it be for the next 3 questions?

Answer 14

It is an upside down parabola. If f(x) increases in the interval (-infinity, -1.5), it decreases in the interval (-1.5, infinity).

Answer 15

ok

Answer 16

c) x = -b/2a = -1.5. That is the vertex of the parabola. Therefore, f(x) attains its maximum at x = -1.5.

Answer 17

d) the function has no relative minima.

Answer 18

i got number 4

Answer 19

That is it.

Answer 20

could you help me in other 4 like this?