1f'(x)=3(λ+2)x2-6λx+9λ
f(x) decreases => f'(x)<0 for all x
=> (λ+2)x2-2λx+3λ<0
now find the solution set of x in the above inequality and then the values of λ can be found.
:)
For what values of λ does the function f (x)=(λ+2)x3−3λx2+9λx−1 decrease for all x?
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11 Answers
1
11yar itna to mujhe bhi aata hai [2]
i am asking 4 an answer as my answer is not matching[1]
11sir ji tussi hamesha di tarah perfect ho
but in the last step
we got
2λ2+6λ>0
so in this what we will take in the interval
λ<-3 and λ>0
or
λ<-3 or λ>0
66There is one more constraint. The coefficient of x2 i.e. λ + 2 < 0. This wont be true for λ>0.
11sir i got
NO SOLUTION
PLEASE CHECK UR WORK AGAIN
BUT AS THE ANSWER GVEN IS NOT MATCHING AS USUAL
PLEASE SHOW ME THE LAST STEPS
66The condition on the derivative give us
(λ+2)x2 - 2λx + 3λ<0
for all x.
For this to happen, the graph of the quadratic f(x)=(λ+2)x2 - 2λx + 3λ must open downwards and be entirely below the x axis. For this we require λ+2<0 (opening downwards) and 4(λ2-3λ(λ+2))<0 (entirely below the axis).
The second condition gives us λ(λ+3)>0 which gives λ>0 OR λ<-3.
So, \lambda \in (-\infty,\,-2)\cap \big((-\infty,\,-3)\cup (0,\,\infty)\big)
which gives λ<-3