Question #273995

For the following functions:

(a) f(x)=(x+2)²(x-1)

(b) f(x)= (x²)/(x²-9)

(c) f(x)= (x²-2x+1)/(x+1)


i. Find the critical numbers off (if any);

ii.Find the open interval(s) on which the function. increasing or decreasing

iii. Apply the first derivative test to identify all relative extrema;

iv. Use graphing utility to confirm your results.


1
Expert's answer
2021-12-05T16:55:15-0500

critical point of a function is point where it function is not differentiable or its derivative is 0


a)

f(x)=2(x+2)(x1)+(x+2)2=3x(x+2)=0f'(x)=2(x+2)(x-1)+(x+2)^2=3x(x+2)=0

f(x)>0f'(x)>0 at x(,2)x\isin (-\infin,-2) - f(x)f(x) is increasing

f(x)<0f'(x)<0 at x(2,0)x\isin (-2,0) - f(x)f(x) is decreasing

f(x)>0f'(x)>0 at x(0,)x\isin (0,\infin) - f(x)f(x) is increasing

f(x)f(x) has relative maximum at x = -2

f(x)f(x) has relative minimum at x = 0


b)

f(x)=2x(x29)2x3(x29)2=18x(x29)2=0f'(x)=\frac{ 2x(x^2-9)-2x^3}{(x^2-9)^2}=-\frac{ 18x}{(x^2-9)^2}=0

function is not differentiable at x=±3x=\pm 3

f(x)>0f'(x)>0 at x(,3)(3,0)x\isin (-\infin,-3)\cup (-3,0) - f(x)f(x) is increasing

f(x)<0f'(x)<0 at x(0,3)(3,)x\isin (0,3)\cup (3,\infin) - f(x)f(x) is decreasing

f(x)f(x) has relative maximum at x = 0


c)

f(x)=x22x+1x+1=(2x2)(x+1)(x1)2(x+1)2=(x1)(x+3)(x+1)2=0f'(x)=\frac{ x^2-2x+1}{x+1}=\frac{ (2x-2)(x+1)-(x-1)^2}{(x+1)^2}=\frac{ (x-1)(x+3)}{(x+1)^2}=0

function is not differentiable at x=1x= -1

f(x)>0f'(x)>0 at x(,3)x\isin (-\infin,-3) - f(x)f(x) is increasing

f(x)<0f'(x)<0 at x(3,1)x\isin (-3,1) - f(x)f(x) is decreasing

f(x)>0f'(x)>0 at x(1,)x\isin (1,\infin) - f(x)f(x) is increasing

f(x)f(x) has relative maximum at x = -3

f(x)f(x) has relative minimum at x = 1




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