a. Find the open intervals on which the function is increasing and decreasing. b

Question

a. Find the open intervals on which the function is increasing and decreasing. b. Identify the function's local and absolute extreme values, if any, saying where they occur. f(x) = x^2 - 8/x - 3, x notequalto 3 a. Find the open intervals on which the function is increasing and decreasing. A. The function is increasing on the subintervals (-infinity, 3) and decreasing on the subintervals (3, infinity) with a discontinuity at x = 3. B. The function is decreasing on the subintervals (-infinity, 2), (4, infinity) and increasing on the subintervals (2, 3), (3, 4) with a discontinuity at x = 3. c. The function is increasing on the subintervals (-infinity, 2) (4, infinity) and decreasing on the subintervals (2, 3), (3, 4) with a discontinuity at x = 3. D. The function is increasing on the subintervals (infinity, 2), (4, infinity) and decreasing on the subintervals (2, 4). b. Identify the function's local extreme values, if any. A. The function has a local maximum at x = 2 and a local minimum at x = 4. B. The function has a local minimum at x = 2 and a local maximum at x = 4. C. The function has no local extrema. D. The function has a local maximum at x = 2 and x = 4 and a local minimum at x = 3. Identify the function's absolute extreme values, if any. A. The function has no absolute extrema. B. The function has an absolute maximum at x = 2 and an absolute minimum at x = 4. C. The function has an absolute maximum at x = 2 and x = 4 and an absolute minimum at x = 3. D. The function has an absolute minimum at x = 2 and an absolute maximum at x = 4

Explanation / Answer

a) we have given f(x)=(x^2-8)/(x-3) where x is not equal to 0

f'(x)=[(x-3)*2x-(x^2-8)]/(x-3)^2=0

2x^2-6x-x^2+8=0

x^2-6x+8=0

x^2-4x-2x+8=0

x(x-4)-2(x-4)=0

(x-4)(x-2)=0

x=2,4

at x=1,f'(1)=[(-2)*2-(-7)]/(-2)^2=3/4>0

at x=5,f'(5)=[(2)*10-(25-8)]/(2)^2=(20-17)/4>0

at x=2.1

f'(2.1)=[(x-3)*2x-(x^2-8)]/(x-3)^2=[(2.1-3)*2(2.1)-((2.1)^2-8)]/(2.1-3)^2<0

at x=3.1

f'(3.1)=[(3.1-3)*2*(3.1)-((3.1)^2-8)]/(3.1-3)^2<0

so the function f(x) is increasing on the subintervals (-infinity ,2), (4,infinity) and decreasing on the subintervals (2,3),(3,4) with discontinuity at x=3

answer is option C

b) At x=2, f'(x) sign changes positive to negative sign, so f(x) has local minimum there.

At x=4,f'(x) sign changes negative to positive sign, so f(x) has local maximum there.

answer is option B

c) We have given f(x)=(x^2-8)/(x-3)

At x=2

f(2)=(4-8)/(2-3)=-4/-1=4

At x=4

f(x)=(16-8)/(1)=8

the function has an absolute minimum at x=2 and an absolue maximum at x=4

answer is option D

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