could you please provide explanations (i) Given that x = 1 is a critical point of the function f(x) = x¹ - ³x² + 2x + 1, find all criticalpoints and characterize them. Sketch a graph of the function. Find also the absolute maxmimumwhen the domain for x is the interval 0 ≤ x ≤ 4.(ii) Draw a contour plot of the function z(x, y) = x² - y² in the xy-plane, and sketch a graph ofthis function in the xyz Cartesian frame.

(i) Given function is Here critical points obtained from Now given that is a critical point of…

Answered: (i) Given that x = 1 is a critical…

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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could you please provide explanations

(i) Given that x = 1 is a critical point of the function f(x) = x¹ - ³x² + 2x + 1, find all critical
points and characterize them. Sketch a graph of the function. Find also the absolute maxmimum
when the domain for x is the interval 0 ≤ x ≤ 4.
(ii) Draw a contour plot of the function z(x, y) = x² - y² in the xy-plane, and sketch a graph of
this function in the xyz Cartesian frame.

Expert Solution

Step 1: optimization of given function

(i) Given function is $f left parenthesis x right parenthesis equals 1 fourth x to the power of 4 minus 3 over 2 x squared plus 2 x plus 1$

Here critical points obtained from $f apostrophe left parenthesis x right parenthesis equals 0 rightwards double arrow x cubed minus 3 x plus 2 equals 0$

Now given that $x equals 1$ is a critical point of given function .

$s o space f apostrophe left parenthesis 1 right parenthesis equals 0 comma space t h e n space f apostrophe left parenthesis x right parenthesis equals 0 rightwards double arrow x squared open parentheses x minus 1 close parentheses plus x open parentheses x minus 1 close parentheses minus 2 open parentheses x minus 1 close parentheses equals 0$ $rightwards double arrow x equals 1 comma 1 comma negative 2$

Now $f apostrophe apostrophe open parentheses x close parentheses equals 3 open parentheses x squared minus 1 close parentheses space a n d space f apostrophe apostrophe apostrophe open parentheses x close parentheses equals 6 x space$

Here at $x equals negative 2 comma f apostrophe apostrophe left parenthesis negative 2 right parenthesis equals 9 greater than 0$ ,so $f space h a s space l o c a l space m i n i m u m space v a l u e space a t space x equals negative 2 space$ .

Now $f apostrophe apostrophe left parenthesis 1 right parenthesis equals 0 space$ ,so from here we can not arise any conclusion about local minima/local maxima .

Now $f to the power of apostrophe apostrophe apostrophe end exponent open parentheses 1 close parentheses equals 6 not equal to 0$ , so at $x equals 1$ , $f$ has neither local minimum nor local maximum value .

Now noted that $f apostrophe open parentheses x close parentheses equals open parentheses x minus 1 close parentheses squared open parentheses x plus 2 close parentheses$

when $0 less or equal than x less than 1 space t h e n space f apostrophe left parenthesis x right parenthesis space greater than 0 space a n d space w h e n space 1 less than x less or equal than 4 space t h e n space f apostrophe left parenthesis x right parenthesis greater than 0$

from below graph we have seen that f is non decreasing function on $open square brackets 0 comma 4 close square brackets$ .

so absolute maximum of f on $open square brackets 0 comma 4 close square brackets$ is obtained at $x equals 4$

which is $f left parenthesis 4 right parenthesis equals 64 minus 24 plus 8 plus 1 equals 73 minus 24 equals 49$