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Home  >>  CBSE XII  >>  Math  >>  Application of Derivatives
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Find the points of local maxima,local minima and the points of inflection of the function $f(x)=x^5-5x^4+5x^3-1$.Also find the corresponding local maximum and local minimum values.

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Toolbox:
  • To obtain the absolute maxima or minima for the function $f(x)$
  • (i) Find $f'(x)$ and put $f'(x)=0$
  • (ii) Obtain the points from $f'(x)=0$
  • (iii) By phthogoras theorem, find $f''(x)$ and check the value of $f''(x)$ for each of the points obtained if $f''(x)>0$ it has a minimum at that point. If $f''(x)<0$ then it has a maximum at that point.
Step 1
$f(x)=x^5-5x^4+5x^3-1$
Differentiating w.r.t $x$ we get,
$f'(x)=5x^4-20x^3+15x^2$
$f'(x)=5x^2(x^2-4x+3)$
when $f'(x)=0$
$ \Rightarrow 5x^2(x^2-4x+3)=0$
$ \Rightarrow 5x^2(x-3)(x-1)=0$
$ \Rightarrow x=0,\: x=3,\: x=1$
Step 2
Difficult $f'(x)$ again w.r.t $x$
$f''(x)=20x^3-60x^2+30x$
when $x=0\: ;\: f''(x)=0$
Step 3
when $x=3$
$ f''(x)=20(3)^3-60(3)^2+30(3)$
$ = 540-540+90$
$ \Rightarrow f''(x)>0$
Hence $x=3$ is a point of local minimum.
Step 4
When $x=1$
$ f''(x)=20(1)^3-60(1)^2+30(1)$
$ = 20-60+30$
$ = -10$
$ \Rightarrow f''(x)<0$
hence $x=1$ is the point of local maximum.
Step 5
Substituting $x=3$ in $f(x)$ we get
$f(3)=3^5-5(3)^4+5(3)^3-1$
$ = 243-405+135-1$
$=-28$
$f(1)=1^5-5(1)^4+5(1)^3-1$
$ = 1-5+5-1$
$ = 0$
Hence local maximum $= 0$
local minimum $ = -28$
$ x = 0$ is the point of inflection.
answered Aug 7, 2013 by thanvigandhi_1
 

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