Questions: Let f(x)=14-5x-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= 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". In the last two, your answer should be a comma separated list of x values or the word "none".

Let f(x)=14-5x-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=
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".

In the last two, your answer should be a comma separated list of x values or the word "none".

Transcript text: (1 point) Let $f(x)=14-5 x-x^{2}$. 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 $\square$ 2. $f$ is decreasing on the intervals $\square$ 3. The relative maxima of $f$ occur at $x=$ $\square$ 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". In the last two, your answer should be a comma separated list of $x$ values or the word "none".

Solution

Solution Steps

Step 1: Find the Derivative of $ f(x) $

To determine where the function $ f(x) = 14 - 5x - x^2 $ is increasing or decreasing, we first find its derivative. The derivative $ f'(x) $ is given by:

\[ f'(x) = \frac{d}{dx}(14 - 5x - x^2) = -5 - 2x \]

Step 2: Determine Critical Points

To find the critical points, we set the derivative equal to zero and solve for $ x $:

\[ -5 - 2x = 0 \]

Solving for $ x $, we get:

\[ 2x = -5 \quad \Rightarrow \quad x = -\frac{5}{2} \]

Step 3: Test Intervals Around Critical Points

We test the intervals around the critical point $ x = -\frac{5}{2} $ to determine where $ f(x) $ is increasing or decreasing.

For $ x < -\frac{5}{2} $, choose $ x = -3 $: \[ f'(-3) = -5 - 2(-3) = -5 + 6 = 1 > 0 \] So, $ f(x) $ is increasing on $(- \infty, -\frac{5}{2})$.
For $ x > -\frac{5}{2} $, choose $ x = 0 $: \[ f'(0) = -5 - 2(0) = -5 < 0 \] So, $ f(x) $ is decreasing on $(- \frac{5}{2}, \infty)$.

Step 4: Determine Relative Maxima and Minima

Since $ f(x) $ changes from increasing to decreasing at $ x = -\frac{5}{2} $, there is a relative maximum at this point.