Does g(x) = xe^kx Have a Relative Extremum at x = -1/k or x = 1/k?

[haris123]

If g(x) = xe^kx where k < 0 is a constant, then we may conclude that g has

(a) a relative maximum at x = -1/k (b) a relative minimum at x = -1/k
(c) a relative maximum at x = 1/k (d) a relative minimum at x = 1/k
(e) none of (a) - (d)

 

[haris123]

i think the answer should be A

 

[Dick]

i think the answer should be A

That would be right. But why do you think so?

 

[haris123]

That would be right. But why do you think so?

the reason is because the graph is decreasing from - infinity to -1/k and increasing from -1/k to infinity. it changes from negative to positive. hence its a maximum. am i right?

 

[Dick]

the reason is because the graph is decreasing from - infinity to -1/k and increasing from -1/k to infinity. it changes from negative to positive. hence its a maximum. am i right?

If that's what you think it does, you are wrong. Changing from decreasing to increasing sounds more like a minimum. Can't you give a reason in terms of derivatives?

 

[haris123]

i am the one who needs help. so you tell me what do you think in terms of derivatives? btw from negative to positive is maximum

 

[zorro]

You can find out if a point is max. or min. by two methods -

1) Check the sign of f'(x) to the left and right of the point. If it's sign is negative to the left and positive to the right, then the point is a local minima.

2) Find out the value of the second derivative at the given point and check its sign. If it is negative then the point is a local maxima. If it is positive then the point is a local minima.

 

[zorro]

btw from negative to positive is maximum

If f'(x) changes sign from negative to positive as we go through a point from left to right, then that point is a minimum not maximum! Review your calculus notes