Applications Of Derivatives Question 339
Question: Maximum slope of the curve $ y=-x^{3}+3x^{2}+9x-27 $ is
[MP PET 2001]
Options:
A) 0
B) 12
C) 16
D) 32
Show Answer
Answer:
Correct Answer: B
Solution:
$ y=f(x)=-x^{3}+3x^{2}+9x-27 $
The slope of this curve $ {f}’(x)=-3x^{2}+6x+9 $
Let $ g(x)={f}’(x)=-3x^{2}+6x+9 $
Differentiate with respect to x, $ {g}’(x)=-6x+6 $
Put $ {g}’(x)=0 $
therefore $ x=1 $ Now, $ {g}’’(x)=-6<0 $ and
Hence at $ x=1, $
$ g(x) $ (slope) will have maximum value. $ {{[g(1)]} _{max\text{.}}}=-3\times 1+6+9=12 $ .
BETA
