answer this follow the format in the example

SOLUTION: Given: f(x) = x3 + 6x2 + 9x Step 1: Compute for the first derivative. f'(x) = 3x2 + 12x + 9 Step 2: Solve for the critical points. Equate f'(x) = 0 and solve for x. f'(x) = 3x2 + 12x + 9 = 0 3x2 + 12x + 9 =0 3(x2 + 4x + 3) = 0 3(x + 1)(x + 3) = 0= (x + 1)(x+ 3) = 0 x+1=0 x+3=0 X =-1 X = -3 Critical Points at x = -1 and at x = -3. Step 3: Plot the critical points on the number line. and note the corresponding intervals. OO > too interval A Interval B interval C (-00, -3) (-3, -1) (-1, +00) Step 4: Analyze the behavior of the graph in each interval. Choose any value of x found in the given interval as a test point. (Test points are randomly chosen found within the interval.) Note that f'(x) = 3x2 + 12x + 9. Interval A: let x = -5 = f'(-5) = 24> 0 = INCREASING Interval B: let x = -2 = f'(-2) = -3 0 = INCREASING Step 5: For horizontal tangents, make use of the critical points found noting that at these points, the value of the derivatives is equal to zero. At x = -3 = f'(-3) = 0 = horizontal tangent at x = -3. At x = -1 => f'(-1) = 0 = horizontal tangent at x = -1. Summary of Answers: Interval A: (-co, -3) = function is INCREASING Interval B: (-3, -1) = function is DECREASING Interval C: (-1, +oo) => function is INCREASING Critical Point: x = -3 = HORIZONTAL TANGENT Critical Point: x =-1 = HORIZONTAL TANGENT -OO +oo interval A Interval B Interval C (-00, -3) (-3, -1) (-1, +00)1. f(x) = 2x -1SOLUTION: Given: f(x) = x2 - 5x + 6 Step 1: Compute for the first derivative. f'(x) = 2x - 5 Step 2: Solve for the critical points. Equate f'(x) = 0 and solve for x. f'(x) = 2x - 5 = 0 2x - 5 = 0 = x = 2.5 Critical Point at x = 2.5. There is only one critical point. Step 3: Plot the critical points on the number line. and note the corresponding intervals. OO +oo 2.5 interval A Interval B (-00, 2.5) (2.5, +00) Step 4: Analyze the behavior of the graph in each interval. Choose any value of x found in the given interval as a test point. (Test points are randomly chosen found within the interval.) Note that f'(x) = 2x - 5. Interval A: let x = 0 = f'(0) = -5 f'(3) = 1 > 0 = INCREASING Step 5: For horizontal tangents, make use of the critical points found noting that at these points, the value of the derivatives is equal to zero. At x = 2.5 = f'(2.5) = 0 => horizontal tangent at x = 2.5. Summary of Answers: Interval A: (-co, 2.5) => function is DECREASING Interval B: (2.5, +00) => function is INCREASING Critical Point: x = 2.5 = HORIZONTAL TANGENT -OO +oo 2.5 Interval A interval B (-00, 2.5) (2.5, +00)