(1 point) Let \( f(x)=x^{3}-9 x^{2}+15 x-7 \). 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". In the last two, your answer should be a comma separated list of \( x \) values or the word "none".

To determine the intervals where the function \( f(x) = x^3 - 9x^2 + 15x - 7 \) is increasing or decreasing, we first need to find the derivative \( f'(x) \) and analyze its critical points. ### Step 1: Find the derivative The derivative of \( f(x) \) is given by: \[ f'(x) = 3x^2 - 18x + 15 \] ### Step 2: Find critical points To find the critical points, we set the derivative equal to zero: \[ 3x^2 - 18x + 15 = 0 \] We can simplify this equation by dividing everything by 3: \[ x^2 - 6x + 5 = 0 \] Now, we can factor the quadratic: \[ (x - 1)(x - 5) = 0 \] Thus, the critical points are: \[ x = 1, \quad x = 5 \] ### Step 3: Test intervals We will test the intervals determined by the critical points \( x = 1 \) and \( x = 5 \): - Interval 1: \( (-\infty, 1) \) - Interval 2: \( (1, 5) \) - Interval 3: \( (5, \infty) \) We will choose test points from each interval: - For \( x = 0 \) in \( (-\infty, 1) \): \[ f'(0) = 3(0)^2 - 18(0) + 15 = 15 > 0 \quad \text{(increasing)} \] - For \( x = 3 \) in \( (1, 5) \): \[ f'(3) = 3(3)^2 - 18(3) + 15 = 27 - 54 + 15 = -12 < 0 \quad \text{(decreasing)} \] - For \( x = 6 \) in \( (5, \infty) \): \[ f'(6) = 3(6)^2 - 18(6) + 15 = 108 - 108 + 15 = 15 > 0 \quad \text{(increasing)} \] ### Step 4: Summarize intervals From our tests, we find: 1. \( f \) is increasing on the intervals: \( (-\infty, 1), (5, \infty) \) 2. \( f \) is decreasing on the intervals: \( (1, 5) \) ### Step 5: Determine relative maxima and minima To find the relative maxima and minima, we evaluate the function at the critical points: - At \( x = 1 \): \[ f(1) = 1^3 - 9(1)^2 + 15(1) - 7 = 1 - 9 + 15 - 7 = 0 \] - At \( x = 5 \): \[ f(5) = 5^3 - 9(5)^2 + 15(5) - 7 = 125 - 225 + 75 - 7 = -32 \] Since \( f \) changes from increasing to decreasing at \( x = 1 \), it is a relative maximum. Since \( f \) changes from decreasing to increasing at \( x = 5 \), it is a relative minimum. ### Final Answers 1. \( f \) is increasing on the intervals: \( (-\infty, 1), (5, \infty) \) 2. \( f \) is decreasing on the intervals: \( (1, 5) \) 3. The relative maxima of \( f \) occur at \( x = 1 \) 4. The relative minima of \( f \) occur at \( x = 5 \)