Test the following functions for extremum:- i. f(x)=2 x^3-15...

Test the following functions for extremum:- i. $f(x)=2 x^{3}-15 x^{2}-84 x+8$. ii. $\quad f(x)=x^{3}-6 x^{2}+9 x-8$. iii. $\quad f(x)=-\frac{3}{4} x^{4}-8 x^{3}-\frac{45}{2} x^{2}+105$. iv. $\quad f(x)=\frac{3}{4} x^{4}-x^{3}-9 x^{2}+7$ v. $\quad f(x)=x^{4}-8 x^{3}+22 x^{2}-24 x+12$. vi. $\quad f(x)=x(x+1)^{3}(x-3)^{2}$ vii. $\quad f(x)=\frac{x^{2}-3 x+2}{x^{2}+2 x+1}$. viii. $f(x)=3 \sqrt[3]{x^{2}}-x^{2}$. ix. $f(x)=\sqrt[3]{(x-1)^{2}}+\sqrt[3]{(x+1)^{2}}$. x. $f(x)=-2 x, \quad x<0$ $=3 x+5, \quad x \geq 0 .\{$ xi. $\begin{aligned} f(x) &=2 x^{2}+3, \quad x \neq 0 \\ &=4, \quad x=0 .\{\text { Ans. maxima at } 0\} \end{aligned}$ $f(x)=\frac{50}{3 x^{4}+8 x^{3}-18 x^{2}+60}$ $f(x)=\sqrt{e^{x^{2}}-1}$ $f(x)=x e^{x}$ $f(x)=x^{4} e^{-x^{2}}$ $f(x)=x^{2} e^{-x}$ xvii. $\quad f(x)=\frac{4 x}{x^{2}+4}$ xviii. $\quad f(x)=-x^{2} \sqrt[5]{(x-2)^{2}}$ xix. $\quad f(x)=\frac{14}{x^{4}-8 x^{2}+2}$ $f(x)=\sqrt[3]{2 x^{3}+3 x^{2}-36 x}$ xx. $\quad f(x)=\sqrt[3]{2 x^{3}+3 x^{2}-36 x}$. xxi. $\quad f(x)=x^{2} \ln x$ xxii. $\quad f(x)=x \ln ^{2} x$ xxiv. $\quad f(x)=|x|+|x-1|+|x-2|$. xxv. $\quad f(x)=\sin ^{4} x+\cos ^{4} x, 0<x<\frac{\pi}{2}$ xxvi. $\quad f(x)=\sin 2 x-x,-\frac{\pi}{2} \leq x \leq \frac{\pi}{2}$. xxvii. $f(x)=\sin x+\frac{1}{2} \cos 2 x, 0 \leq x \leq \frac{\pi}{2}$.

Test the following functions for extremum:-
i. $f(x)=2 x^{3}-15 x^{2}-84 x+8$.
ii. $\quad f(x)=x^{3}-6 x^{2}+9 x-8$.
iii. $\quad f(x)=-\frac{3}{4} x^{4}-8 x^{3}-\frac{45}{2} x^{2}+105$.
iv. $\quad f(x)=\frac{3}{4} x^{4}-x^{3}-9 x^{2}+7$
v. $\quad f(x)=x^{4}-8 x^{3}+22 x^{2}-24 x+12$.
vi. $\quad f(x)=x(x+1)^{3}(x-3)^{2}$
vii. $\quad f(x)=\frac{x^{2}-3 x+2}{x^{2}+2 x+1}$.
viii. $f(x)=3 \sqrt[3]{x^{2}}-x^{2}$.
ix. $f(x)=\sqrt[3]{(x-1)^{2}}+\sqrt[3]{(x+1)^{2}}$.
x. $f(x)=-2 x, \quad x<0$
$=3 x+5, \quad x \geq 0 .\{$
xi. $\begin{aligned} f(x) &=2 x^{2}+3, \quad x \neq 0 \\ &=4, \quad x=0 .\{\text { Ans. maxima at } 0\} \end{aligned}$
$f(x)=\frac{50}{3 x^{4}+8 x^{3}-18 x^{2}+60}$
$f(x)=\sqrt{e^{x^{2}}-1}$
$f(x)=x e^{x}$
$f(x)=x^{4} e^{-x^{2}}$
$f(x)=x^{2} e^{-x}$
xvii. $\quad f(x)=\frac{4 x}{x^{2}+4}$
xviii. $\quad f(x)=-x^{2} \sqrt[5]{(x-2)^{2}}$
xix. $\quad f(x)=\frac{14}{x^{4}-8 x^{2}+2}$
$f(x)=\sqrt[3]{2 x^{3}+3 x^{2}-36 x}$
xx. $\quad f(x)=\sqrt[3]{2 x^{3}+3 x^{2}-36 x}$.
xxi. $\quad f(x)=x^{2} \ln x$
xxii. $\quad f(x)=x \ln ^{2} x$
xxiv. $\quad f(x)=|x|+|x-1|+|x-2|$.
xxv. $\quad f(x)=\sin ^{4} x+\cos ^{4} x, 0<x<\frac{\pi}{2}$
xxvi. $\quad f(x)=\sin 2 x-x,-\frac{\pi}{2} \leq x \leq \frac{\pi}{2}$.
xxvii. $f(x)=\sin x+\frac{1}{2} \cos 2 x, 0 \leq x \leq \frac{\pi}{2}$.

Key Concepts

Piecewise Functions

Piecewise functions are defined by different expressions over different intervals of the domain. Each segment may have its own set of critical points and potential non-differentiable junctions. Analyzing these functions involves a careful examination of each piece, particularly at the boundaries where the definition changes, to ensure that all potential extrema are properly identified.

Extremum and Extreme Values

Extremum refers to the highest (maximum) or lowest (minimum) values a function attains within a given domain. These can be categorized as local extrema, which are the highest or lowest values in a small neighborhood, and global extrema, which are the absolute highest or lowest values over the entire domain. This concept is fundamental in optimization and analysis of functions.

Critical Points

Critical points are points in the function's domain where the derivative is zero or does not exist. These points are potential candidates for being local extrema and are the starting point of any analysis aimed at determining the maximum or minimum values of a function.

First Derivative Test

The first derivative test involves studying the sign of the derivative on either side of a critical point. If the derivative changes from positive to negative, the critical point is a local maximum; if it changes from negative to positive, it's a local minimum. This test is widely used because it directly links the rate of change in the function to its increasing or decreasing behavior.

Second Derivative Test

The second derivative test provides additional information by examining the concavity of the function at a critical point. A positive second derivative indicates that the function is concave up, suggesting a local minimum, while a negative second derivative indicates concavity down, suggesting a local maximum. This test is particularly effective when the first derivative is zero.

Differentiability and Points of Non-differentiability

Differentiability is a key concept in extremum analysis, as derivative-based tests require the function to be differentiable. However, functions can have points where they are not differentiable (such as cusps, corners, or points defined by absolute value or root functions), and these points can also be candidates for extrema. Recognizing and analyzing such points often require methods beyond standard derivative tests.

Domain and Boundary Analysis

The domain of a function is critical in determining its extrema, as global extrema need to be considered over the entire valid input interval. In cases where the domain is closed or limited, the endpoints must be evaluated separately since they might yield absolute extrema even if they are not critical points in the classical sense.

Transcript

00:01 The first function will determine the following function here.

00:03 The first function is given as the mode of x negative 1.

00:05 It's given as, come out of having x negative 1 or x negative 1 greater than equal to 0 or x is greater than 0 or negative of that negative x negative 1 is equal to 1 negative x or we have x negative 1 less than 0 or x less than 1.

00:28 The next function is given as mode of 2x positive 1.

00:32 That is given as 2x positive 1 for 2x plus 1 greater than equal to 0 or x is greater than equal to negative 1 over it.

00:40 Or we have negative 2x negative negative 1 over 2.

00:48 Next function is given as mod of lnx that is equal to we have positive allanx for lmx squared as then or equal to 0.

00:58 So by the graph of here we have ln x so lnx is greater than equal to 0.

01:02 This part here we have lnx is greater than equal to 0 x x is lnx greater than equal to 0 we have so this is possible for x is greater than equal to 1 so for x is greater than equal to 1 or this is given as negative lnx for next we have negative alance we have this part here is for 0 less than x less than 1 next we have given as e to the power x take any function if the part x is always positive me have.

01:44 Always positive, it is a part x only.

01:47 The motor way to the part x is if the part x.

01:48 Always positive.

01:52 Next, signum, x plus 3.

01:54 That is given...

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