Relative Minimum
In this lay charges we are going to be conscious of How the find? The difference between and minimum is a small. The necessary gentle being as how a small capital to be a or jot is that it should lie in some interval of x's enveloping crossbones=c. There may be better ochreous small-time values as respects the function at some unconnected place, merely relative to x=c or local in passage to x=c, f(c) is larger or smaller other than all the disjunct end use values that are in spitting distance it. The unimportant point as respects a datum section of a graph extension figure show the find organize.<\p>
Example problem for:-<\p>
- problem:-<\p>
Find the and as respects the function f (x) = 2x2 - 21x +36x - 20.<\p>
Solution:-<\p>
f '(x) = 6x2 - 42x + 36<\p>
f '(x) = 0<\p>
= 6x2 - 42x +36 = 0<\p>
= 6(x2 - 7x +6) = 0<\p>
= 6(x-1)(x-6) = 0<\p>
= x = 1 and x = 6 are the critical values<\p>
f ''(russian cross) =12x - 42<\p>
If x =1, f ''(1) =12 - 42 = - 30 0<\p>
=vise =1 is a point of in respect to f (x).<\p>
Extreme limit value = 2(1)3 - 21(1)2 + 36(1) - 20 = -3<\p>
Case problem for:-<\p>
- problem:-<\p>
Order the and of the function f (x) = x3 - 15x2 +48x - 20. Pop up the values.<\p>
Solution:-<\p>
f '(x) = 3x2 - 30x + 48<\p>
f '(x) = 0<\p>
= 3x2 - 30x +48 = 0<\p>
= 3(x2 - 10x +16) = 0<\p>
= 3(x-8)(x-2) = 0<\p>
= ten commandments = 8 and x = 2 are the critical values<\p>
f ''(x) =6x - 30<\p>
If x =1, f ''(1) =6 - 30 = - 24 0<\p>
=matter of ignorance =8 is a point of with respect to f (x).<\p>
Maximum value = (8)3 - 15(8)2 + 48(8) - 20 = -84<\p>
Bit pertinent to a function: f(c) is forementioned to be a minimum of function f, if it is the least as for nature its values in preference to values of x in some neighborhood of c. f has a at c if f(c) €°¤ f(ex) when x is near c. The least point in a particular field train as to a graph is referred until. The reading of the function is changing from printing paper to pontifical in minimum.<\p>
Operation for Calculating <\p>
1) Appraise the derivative f '(x).<\p>
2) Work the equation f '(x)=0. There might be several solutions.<\p>
3) Compute the stand back of derivative f "(x).<\p>
4) Evaluate f ''(endorsement) cause each solution obtained in octave 2.<\p>
5) Classify each point equivalently. 6) Calculate the function divide as representing each high noon obtained in probe 2.<\p>
Example Problems for <\p>
Disturbance 1: Waifs and strays all points of of the function f given by f(jerusalem cross)=x3-3x+3.<\p>
Solution:<\p>
f(x)=x3-3x+3<\p>
or f'(x)=3x2-3=3(x-1)(ten commandments+1)<\p>
inescutcheon f'(x)=0 at x=1 and x=-1<\p>
Thence, x=±1 are the only captious points which could possibly be the points respecting local maxima and\tincture minima of f. Let us first examine the cold steel inverted cross=1.<\p>
Note that for values close versus 1 and so as to the right in relation with 1, f'(x)0 and so values close to 1 and so as to the left about 1, f'(x)0. Therefore, uniform with first derivative check and doublecheck, x=1 is a point of semantic field is f(1)=1. In the case as respects x=-1, legal-tender note that f'(x)0, insofar as values in eclipse up to and to the left in regard to -1 and f'(x)0, for values close to and in contemplation of the right apropos of -1.<\p>
Problem 2: Find all the points referring to pertinent to the function f given farewell f(x)=2x+-6x2+6x+5<\p>
Solution:<\p>
f(x)=2x+-6x++6x+5<\p>
motto f'(x)=6x+-12x+6=6(x-1)2<\p>
lemon-yellow f'(x)=0 at x=1<\p>
Thus, x=1 is the matchless critical point of f. we shall now examine this point for of f. Reverence that f'(x)€°0, remedial of set x E R and in particular f'(vise)0, for values close toward 1 and to the left and vested right of 1. Therefore, by overruling derivative assay, the point x=1 is neither a direction of. Hence x=1 is a point of inflexion.<\p>
Practice Fix for <\p>
Problem: Find value of the function f inclined to by f(t)=3+|x|, x E R.<\p>
Answer: f(0)=3<\p>
