logo

Ask Questions, Get Answers

 
X
 Search
Want to ask us a question? Click here
Browse Questions
Ad
Home  >>  CBSE XII  >>  Math  >>  Application of Derivatives
0 votes

At what point,the slope of the curve $y=-x^3+3x^2+9x-27$ is maximum?Also find the maximum slope.

Can you answer this question?
 
 

1 Answer

0 votes
Toolbox:
  • Let $f(x)$ be a function with domain $D\: \subset \: R$. Then $f(x)$ is said to attain the maximum value at a point, $ a \in D,$ if $ f'(x) \leq f(a)$ for all $ x \in D$.
Step 1
$f(r)=-x^3+3x^2+9x-27$
Differentiating w.r.t $x$ we get
$f'(x)=-3x^2+6x+9$
If $f'(x)=0$
$ \Rightarrow -3x^2+6x+9=0$
$ 3(-x^2+2x+3)=0$
factorising this we get,
$ f'(x)=(x+3)(x-1)$
when $x=-3$
$f(x)=-(3)^3+3(3)^2+9(3)-27$
$ = -27+27+27-27$
$=0$
when $ x=+1$
$f(1)=-(+1)^3+3(+1)^2+9(+1)-27$
$ = -16$
Hence the point is $(1, -16)$
The maximum slope =
$ f'(1)=-3(1)+6(1)+9$
$ = 12$
answered Aug 6, 2013 by thanvigandhi_1
 

Related questions

Ask Question
student study plans
x
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App
...