Question

given \[f(x)=k*(k-1)^2*e^{-x}+k\]

a) For what values of x in terms of k will f(x) have exactly 2 stationary points.

b) Find the exact co-ordinates of local max point.

Ok. So I believe we differentiate f(x) and make f'(x)=0 and solve for k. I got k=0 and k=1. But I don't know if its right or not.
Can anyone help me out.

Thanks.

Answer 1

I can help with graph if you did not try that way.

Answer 2

Feel free to try www.calorful.com and manipulate your constant k as you wish.

Answer 3

Thanks for the reply, but could you check my way is correct though, because in the next question, it asks to find the local max and if the graph is a straight line, there is no max.

Answer 4

Correct!
No max for a straight line.
Let me check your way.

Answer 5

So what are your steps?
could you share them with me?

Answer 6

I differentiate f(x) and then make it equal to 0 to solve for k. So I got k=0 and k=1. But i think you sub it back into the equation, however it does not makes sense.

Answer 7

They're asking for values of x though,
not values of k.

Hmm this is weird :p

Answer 8

I have done what you said, but I got nothing!

Answer 9

Like.. umm.. if x=ln(k), then,\[\large\rm f'=-k(k-1)^2e^{-\ln k}\]\[\large\rm f'=-k(k-1)^2e^{\ln \frac1k}\]\[\large\rm f'=-k(k-1)^2\frac1k\]\[\large\rm f'=(k-1)^2\]

Answer 10

Honestly I don't know if that's what they're looking for though...
this problem doesn't make much sense to me :(

Answer 11

Woops that last line should be,\[\large\rm f'=-(k-1)^2\]

Answer 12

why \[x=\ln(k)\]?

Answer 13

i have\[-k*(k-1)^2*e^{-x}\]

Answer 14

yes, e^(-x) not e^(-ln(k))