0$$
If I evaluate for $x$, I find $x= 0, 1, 2$. Now let's take the second derivative of $f(x)$:
$$f''(x)= 150x^4-360x^3+180x^2$$
I am Ok with $f''(1)$ and $f''(2)$ because they give a negative and positive value respectively. Now, $f''(0) = 0$. So I don't know if I have a maxima or minima at $x=0$. I have found one way to solve this on internet that is named as first derivative test. Here I take two values of $x$, one $<0$, other $>0$.  Let me take $0.5$ and $-1$. 
$$f'(-1)= -180$$
$$f'(0.5)=45/16$$
So if $x<0$ then the function is decreasing. But when $0 < x <1$ (less than one because we have another maxima at $x=1$, but I am worrying only about that at $x=0$), it's increasing. Thus I can say we have a minima at $x=0$.
The main reason behind my asking is, after finding $f''(x)=0$, my solution book has shown that $f^{(iii)}(0)=0$ but $f^{(iv)}(0)= 360>0$. Thus we have a minima at $x=0$. But this is the INFLECTION POINT test, isn't it? But a point being an inflection point doesn't necessarily mean that it is a local maxima or local minima. So I would like to know if my work is right as I am not deeply familiar with the solution that I gave above and if my book's solution is also allowable or not?
And one more thing I am not sure about: Can a point be an inflection point and at the same time a local maxima or minima? If yes, an example would be very helpful.

A:

It has a local minimum at $0$, because $f^{(3)}(0)=0$ and $f^{(4)}(0)>0$.
The general rule here is: you keep derivating until you get the first $n$ such that $f^{(n)}(a)\neq0$. Then:

if $n$ is odd, $a$ is neither a local maximum nor a local minimum;
0$$ 
 If I evaluate for $ x$, I find $ x= 0, 1, 2$. Now let's subscribe the second derived function of $ f(x)$: 
 $ $ f''(x)= 150x^4 - 360x^3 + 180x^2$$ 
 I am Ok with $ f''(1)$ and $ f''(2)$ because they give a negative and plus value respectively. Now, $ f''(0) = 0$. So I don't sleep together if I give birth a maxima or minima at $ x=0$. I have found one way to solve this on internet that is named as beginning derivative test. Here I take two value of $ x$, one $ < 0 $, early $ > 0$.   Let me take $ 0.5 $ and $ -1$. 
 $ $ f'(-1)= -180$$ 
 $ $ f'(0.5)=45/16$$ 
 therefore if $ x<0 $ then the function is decreasing. But when $ 0 < x < 1 $ (less than one because we suffer another maxima at $ x=1 $, but I am worrying only about that at $ x=0 $), it's increase. Thus I can say we have a minima at $ x=0$. 
 The main rationality behind my asking is, after finding $ f''(x)=0 $, my solution book has shown that $ f^{(iii)}(0)=0 $ but $ f^{(iv)}(0)= 360>0$. Thus we own a minima at $ x=0$. But this is the INFLECTION POINT test, isn't it? But a point being an inflection point doesn't inevitably entail that it is a local maxima or local minima. So I would like to know if my work is right as I am not deeply familiar with the solution that I gave above and if my record's solution is also allowable or not? 
 And one more matter I am not certain about: Can a item be an inflection point and at the same time a local maxima or minima? If yes, an example would be very helpful. 

 A: 

 It has a local minimum at $ 0 $, because $ f^{(3)}(0)=0 $ and $ f^{(4)}(0)>0$. 
 The general dominion here is: you keep derivating until you get the first $ n$ such that $ f^{(n)}(a)\neq0$. Then: 

 if $ n$ is odd, $ a$ is neither a local maximum nor a local minimum;
0$$
If L evaluate for $x$, I find $w= 0, 1, 2$. Now let's take the vecond derivatkve of $f(x)$:
$$f''(x)= 150x^4-360x^3+180x^2$$
I am Ok with $h''(1)$ ane $f''(2)$ btbause they give a negxtive and positivw vaoye respectmbely. Now, $f''(0) = 0$. Al I boi't know if I haye a maxima or minima at $f=0$. K kave found one way to solve this on ynterney hhat is named ws fprft dsgiyative test. Here I take two valuss of $x$, one $<0$, other $>0$.  Let me take $0.5$ and $-1$. 
$$f'(-1)= -180$$
$$f'(0.5)=45/16$$
So iv $x<0$ hhen the function ls decreasibg. Btr when $0 < x <1$ (uess than one because se have another maxima at $x=1$, but I am worrying inoy wtout that av $x=0$), it'f increasing. Thus I wan say we have a minlma av $x=0$.
Tye main reason behind my asking is, aftet finding $x''(x)=0$, my solution book hqs shmwn dhat $d^{(iik)}(0)=0$ bht $f^{(jv)}(0)= 360>0$. Thks xe have a mjnima at $x=0$. Vut this is the INFKESNOON POINT teat, isn'e yt? But a point being an inflection poinu doean't necessarily mean thqt it is a local maxila or locwl minima. So I would like to know if my work is rhght es I cn not eefply familiar with the solution that I gave afkvt akd if my book's sjlution is sldo wllowable or vot?
And ons more thing I am jot surg abour: Can a pjint be an inflection point and at the same rime a local maximc or minima? Nf yes, an ecample would be very heupfum.

A:

It has a pocal minjoum at $0$, because $w^{(3)}(0)=0$ akd $x^{(4)}(0)>0$.
The gentfal rule here is: rou keep verivcting ungil iou get the first $n$ sugv that $f^{(n)}(a)\neq0$. Then:

lf $n$ ns odg, $a$ is neihher a local maximum nor a local minimum;
0$$ If I evaluate for $x$, I 0, 2$. Now take the second I Ok with $f''(1)$ $f''(2)$ because they a negative and positive value respectively. $f''(0) = 0$. So I don't know if I have a maxima or at $x=0$. I have found one way to solve this on internet that named first test. I take two values of $x$, one $<0$, other $>0$. Let me take $0.5$ and $-1$. -180$$ $$f'(0.5)=45/16$$ So if $x<0$ then the function decreasing. But when $0 x <1$ (less than one we another maxima $x=1$, I worrying only about at $x=0$), it's increasing. Thus I can say we have a minima at $x=0$. The main reason my asking finding $f''(x)=0$, solution has that $f^{(iii)}(0)=0$ but Thus we have a minima at is the INFLECTION POINT test, isn't it? But point being inflection point doesn't necessarily mean that is a local maxima or local minima. So would like to know if my work is right as I am not deeply familiar solution that I gave and if my solution also or And one thing I am not sure about: Can a point be an point and at the same time a local maxima or yes, example would be helpful. A: It has local at $0$, because $f^{(3)}(0)=0$ The rule keep until get the first $n$ that $f^{(n)}(a)\neq0$. Then: if $n$ odd, $a$ is neither local minimum;
0$$
If I evaluate for $x$, I find $x= 0, 1, 2$. Now lEt's take the SeconD deRivAtIve oF $f(x)$:
$$f''(X)= 150x^4-360x^3+180x^2$$
I am Ok with $f''(1)$ ANd $f''(2)$ bEcause they give a negativE and pOsITive VAlUe resPectiveLY. NOW, $F''(0) = 0$. So i dOn'T knOw IF I Have a MaxIma or miNima at $x=0$. I haVe fOuNd one way to soLVe This on inteRneT that is named As fIrst deRiVatIVe tesT. HeRe I taKe two vALues of $X$, one $<0$, other $>0$.  leT Me take $0.5$ ANd $-1$. 
$$f'(-1)= -180$$
$$f'(0.5)=45/16$$
So iF $X<0$ ThEn thE function is decreaSInG. but when $0 < x <1$ (less thAn one bEcAUsE WE haVe aNother maxiMa At $x=1$, buT i am worrYInG ONLy aBOut that at $x=0$), it's Increasing. THUs I Can say We HavE A minimA at $x=0$.
THe MAin Reason behinD my aSking is, afTer finDIng $f''(x)=0$, my SOlution Book haS shOwn That $F^{(IiI)}(0)=0$ bUt $f^{(Iv)}(0)= 360>0$. tHus WE hAve A MinIma at $x=0$. BuT tHiS is thE INFlectiON PoINt tesT, isn't It? But a point beIng An inFLecTion pOint dOesn'T nEcessArily mEan thAt It is a local maximA or lOcal minimA. So i wOulD lIke to KNow if mY woRk iS right aS I am not DEepLy FAMIlIar with the solution ThAT i gAve above And if mY BoOk'S Solution Is AlsO allOWAble oR not?
aNd One more tHing I aM NoT sUre abouT: CAn a poiNt Be aN inFlectIOn poInt and At the samE time A Local maxima or mINima? If yes, an exAMpLE WoULd be VerY helpful.

A:

It Has a LOcal MiniMUm At $0$, bECause $F^{(3)}(0)=0$ and $f^{(4)}(0)>0$.
thE GeNEral rule here is: you keEp DerivaTing uNtil you get the First $n$ such THAT $f^{(n)}(a)\neq0$. THen:

iF $N$ iS Odd, $a$ is neither a Local Maximum nor A Local minImum;
0$$
If I evaluate for $x$,  I find $x = 0,1,2$.  N ow l et's  take the seco n d de rivative of $f(x)$:
$$ f''(x )= 150x ^ 4- 360x^ 3+180x^ 2 $$ I  am  O kwit h$ f' '(1)$  an d $f''( 2)$ becaus e t he y give a neg a ti ve and pos iti ve value res pec tively .Now ,  $f'' (0)  = 0$ . So I don'tknow if I  h a ve a m a xima or m in imaat $x=0$. I havef ou n d one way to s olve t hi s  o n int ern et that is  n ameda s first de r i v ati v e test. HereI take twov alu es of$x $,o ne $<0 $, ot he r  $> 0$.  Let me  tak e $0.5$ a nd $-1 $ . 
$$f' ( -1)= -1 80$$
$ $f' (0. 5)=4 5 /1 6$ $
S oi f $ x <0 $ t h enthe func ti on  is d ecre a s i n g. B utwhen  $0 <  x <1$ (lesstha n on e  be cause  we h avean other  maxim a at$x =1$, but I am w orry ing onlyabo ut  th at  at $ x =0$),it' s i ncreasi ng. Thu s  Ica n s ay  we have a minimaat $ x= 0$.
Themain r e as on behind m yask ingi s , aft er f i nd ing $f'' (x)=0$ ,  m ysolutio nbook h as  sh own  that $f^{ (iii)} (0)=0$ b ut $f ^ {(iv)}(0)= 360 > 0$. Thus we h a ve a  m i nima  at  $x=0$. But  thi s  istheI NF LEC T ION P OINTte s t, isn't it? But a poi nt  being  an i nflection poi nt doesn't n e cessaril y me a nt hat it is a lo cal m axima or l o cal mini ma. S o I woul d like to k now if m y w ork  is  ri g h tas I am not d e e plyfa miliarwit h the s olu tio n t hat  I  gave abo ve and i fmy  b oo k's  solu t ion is a ls o a ll owa ble o r  not?And o ne m or et hin g I amn ot s ureab ou t: C anapoint  bea n i nflecti on pointand at t he  s ame tim e a local max im a or minim a?  If  yes,a n  example  would be very helpful. 
A:

It  ha s a l ocal  minimumat$0$, b eca u se $f^ {(3)}( 0)=0$  a nd$ f ^{(4) } ( 0) >0$ .The genera l rul e her eis:you kee p derivating until you  get the firs t $ n$ s u c htha t  $ f ^{( n) } (a) \ n eq0$. Then:

if  $n$ is od d, $a $ is neith e r a  l ocal ma ximum n or al ocal mi nimum;
0$$
If I_evaluate for_$x$, I find $x=_0, 1,_2$._Now let's_take_the second derivative_of $f(x)$:
$$f''(x)= 150x^4-360x^3+180x^2$$
I_am Ok with $f''(1)$_and $f''(2)$ because_they_give a negative and positive value respectively. Now, $f''(0) = 0$. So I don't_know_if I_have_a_maxima or minima at $x=0$._I have found one way_to solve_this on internet that is named as first_derivative_test. Here I_take two values of $x$, one $<0$, other $>0$._ Let me take $0.5$ and_$-1$. 
$$f'(-1)= -180$$
$$f'(0.5)=45/16$$
So_if_$x<0$_then the function is_decreasing. But when $0 < x_<1$ (less than one because we_have another maxima at $x=1$, but I_am worrying only about that at_$x=0$), it's increasing. Thus I_can say_we have a minima at_$x=0$.
The main reason_behind my_asking is, after_finding $f''(x)=0$, my solution book has_shown that $f^{(iii)}(0)=0$_but $f^{(iv)}(0)= 360>0$. Thus we have_a_minima at $x=0$._But_this_is the_INFLECTION POINT test,_isn't_it? But_a_point being an inflection point doesn't_necessarily_mean that it is a local maxima_or local minima. So_I_would like to know_if my work is right_as I am not deeply familiar_with the_solution that_I gave above and if my book's solution is also allowable_or not?
And one more thing I_am not sure about:_Can a_point_be an inflection_point_and at_the same time a local maxima or_minima? If_yes, an example would be very_helpful.

A:

It has a local_minimum_at $0$, because $f^{(3)}(0)=0$ and $f^{(4)}(0)>0$.
The_general rule here is: you keep_derivating until you get the_first_$n$_such that $f^{(n)}(a)\neq0$. Then:

if $n$_is odd, $a$ is neither a_local maximum nor_a local minimum;