Answer to Question #273995 in Calculus for Josa Kyme Palma

Question

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.

Answer 1

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

a)

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

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

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

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

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

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

b)

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

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

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

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

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

c)

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

function is not differentiable at $x= -1$

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

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

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

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

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