Let f(x)=2x+6/x Find the open intervals on which ff is increasing (decreasing). Then determine the x-coordinates of all relative maxima (minima). **Transcript for Educational Website:**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".

Given: The function f(x)=2x+6x 1. Find on which interval the function is increasing 2. Find on which interval the function is decreasing 3. Find the relative maxima of f occur at x value 4. Find the relative minima of f occur at x value Use the critical points to form the test interval Differentiate the function f(x)=2x+6x with respect to x f(x)=2x+6xf'(x)=2x+6x-1=2+6(-1)x-1-1=2-6x-2f'(x)=2-6x2 Find the critical point solve f'(x)=0 f'(x)=2-6x2f'(x)=02-6x2=02=6x22x2=6x2=62x2=3x=±3 The critical points are x=3 and -31) The test intervals for the critical points are -∞,-3, -3,0, 0,3, and 3,∞ Determine the sign of f'x at one test value in each interval The test value for -∞,-3 is -2 f'(x)=2-6x2f'(-2)=2-6(-2)2=2-64f'(-2)=12 f'(-2)>0 therefore f(x)is increasing on -∞,-3 The test value for 3,∞ is 2 f'(x)=2-6x2f'(2)=2-622=2-64f'(2)=12 f'(2)>0 therefore f(x)is increasing on 3,∞ Answer: f(x) is increasing on the interval -∞,-3, 3, ∞ 2) Determine the sign of f'x The test value for -3,0 is -1 f'(x)=2-6x2f'(-1)=2-6(-1)2f'(-1)=-4 f'(-1)<0 therefore f(x)is decreasing on the interval -3,0 The test value for 0,3 is 1 f'(x)=2-6x2f'(1)=2-612f'(1)=-4 f'(1)<0 therefore f(x)is decreasing on the interval 0,3 Answer: f(x) is decreasing on the interval -3,0, 0,3 3) First derivative test of relative maxima: If f'(x) is positive on a,c and negative on c,b then f has relative maxima at x=c. f'(x) is positive on -∞,-3 and negative on -3,0 then fhas relative maxima at x=-3. The relative maxima of f occur at x=-3.4) First derivative test of relative minima: If f'(x) is negative on a,c and positive on c,b then f has relative minima at x=c. f'(x) is negative on 0,3 and positive on 3,∞ then f has relative minima at x=3. The relative minima of f occur at x=3.