Question

For the following piecewise-defined function f. find the critical numbers, local extreme values, and absolute extreme values on the closed interval 6, 90 20r109 if 6<214 f(z) 14< <18 8 87 if 18 < z<90 10+237 if Critical number(s) Preview Local minimum value(s) Preview Local maximum value(s) Preview Absolute minimum value: Preview Absolute maximum valug: Preview Points possible: 1 This is attempt 1 of 2. Submit

Answer 1

Some formulas using here are

• 
  d(constant) 0 dz
  

• 
  -(constant x f(r) constant x f (x)] dr dr
  

• 
  f()g(r) d(f(x)) d(g() da da dz
  

• 
  nz-1 dz
  

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220r109 if 6 r<14 if 14< r18 if 18< r 90 (a) 8x-87 -10237

Finding critical points by solving $f{}'\left ( x \right )=0$ or not defined.

f (r

For

6 xK 1 4

f (x) -20r109

() -20+109] 2-20r109] dr

$=\frac{\mathrm{d} }{\mathrm{d} x} \left [x^{2} \right ]-\frac{\mathrm{d} }{\mathrm{d} x} \left [20x \right ]+\frac{\mathrm{d} }{\mathrm{d} x} \left [109 \right ]$

2ar-20 x 는 [띠] +0 dr [x] +0

2-20 x

$=2x-20$

For $14< x\leq 18$

$f\left ( x \right )=8x-87$

$f{}'\left ( x \right )=\frac{\mathrm{d} }{\mathrm{d} x}\left [8x-87 \right ]$

$=\frac{\mathrm{d} }{\mathrm{d} x}\left [8x \right ]-\frac{\mathrm{d} }{\mathrm{d} x}\left [87 \right ]$

$=8-0$

$=8$

For $18< x\leq 90$

$f\left ( x \right )=-10x+237$

f(a) -10+237] |-10x 237]

$=\frac{\mathrm{d} }{\mathrm{d} x}\left [-10x \right ]+\frac{\mathrm{d} }{\mathrm{d} x}\left [237 \right ]$

$=-10+0$

$=-10$

That is

$f{}'\left ( x \right )=\left\{\begin{matrix} 2x-20 & if &6\leq x\leq 14 \\ 8 & if &14< x\leq 18 \\ -10 & if & 18< x\leq 90 \end{matrix}\right.$

$f{}'\left ( x \right )=0$

$2x-20=0$

2=20

$x=\frac{20}{2}=10$

To exist a derivative at a point , right hand and left hand derivative at that point should be equal.

At $x=14$

$\lim_{x\rightarrow 14^{-}}f{}'\left ( x \right )=\lim_{x\rightarrow 14}\left ( 2x-20 \right )$

$=2\times 14-20$

$=8$

$\lim_{x\rightarrow 14^{+}}f{}'\left ( x \right )=8$

That is

$\lim_{x\rightarrow 14^{-}}f{}'\left ( x \right )=\lim_{x\rightarrow 14^{+}}f{}'\left ( x \right )=8$

Therefore

$f{}'\left ( 14 \right )=8$

$\lim_{x\rightarrow 18^{-}}f{}'\left ( x \right )=8$

lim f'(x) 18+ OI-

$\lim_{x\rightarrow 18^{+}}f{}'\left ( x \right )\neq \lim_{x\rightarrow 18^{-}}f{}'\left ( x \right )$

Therefore

Derivative does not exist at $x=18$

That is

Critical numbers are

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$f{}'\left ( x \right )=\left\{\begin{matrix} 2x-20 & if &6\leq x\leq 14 \\ 8 & if &14< x\leq 18 \\ -10 & if & 18< x\leq 90 \end{matrix}\right.$

$f{}''\left ( x \right )=\frac{\mathrm{d} }{\mathrm{d} x}\left [ f{}'\left ( x \right ) \right ]$

For

6 xK 1 4

$f{}'\left ( x \right )= 2x-20$

$f{}''\left ( x \right )= \frac{\mathrm{d} }{\mathrm{d} x}\left [2x-20 \right ]$

$= \frac{\mathrm{d} }{\mathrm{d} x}\left [2x \right ]- \frac{\mathrm{d} }{\mathrm{d} x}\left [20 \right ]$

$=2-0$

For $14< x\leq 18$

$f{}'\left ( x \right )=8$

$f{}''\left ( x \right )=\frac{\mathrm{d} }{\mathrm{d} x}\left [8 \right ]=0$

For $18< x\leq 90$

f (r= -10

$f''\left ( x \right )=\frac{\mathrm{d} }{\mathrm{d} x}\left [-10 \right ]=0$

At $x=10$

$f{}''\left ( x \right )= 2> 0$

Function has local minimum at $x=10$

220r109 if 6 r<14 if 14< r18 if 18< r 90 (a) 8x-87 -10237

$f\left ( 10 \right )= 10^{2}-20\times 10+109 =9$

That is

Local minimum value $\boldsymbol{ =9}$

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From above calculations , there is no x values satisfying $f{}''\left ( x \right )< 0$

Therefore

Function does not have local maximum.

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Critical numbers are

For absolute maximum and absolute minimum values , we have to check values of function at boundary points and critical

numbers.

$x=6,10,14,18,90$

220r109 if 6 r<14 if 14< r18 if 18< r 90 (a) 8x-87 -10237

$f\left ( 6 \right )=6^{2}-20\times 6+109 =25$

$f\left ( 10 \right )= 10^{2}-20\times 10+109 =9$

$f\left ( 14 \right )= 14^{2}-20\times 14+109 =25$

$f\left ( 18 \right )=8\times 18-87 =57$

$f\left ( 90 \right )=-10\times 90+237=-663$

That is

Absolute minimum value $\boldsymbol{=-663}$

Absolute maximum value
5

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