Showing increasing function? (1 Viewer)

[rumbleroar]

Q28, I know f'(x)>0 but when I solve for the discriminant, do I also let the discriminant >0? Or is my approach wrong? I got k<3 but I think I should be getting two values for k?

Please clarify!! Thanks in advance ))


Sent from my iPhone using Tapatalk

 

[Carrotsticks]

We want it to be an increasing function, so as you said we must have f'(x)>0.

We now have a quadratic in terms of K, and this quadratic is f'(x). We want this quadratic to be positive (meaning it is concave up and lies ABOVE the x axis). What I mean by a quadratic being positive is that all the Y values are positive.

Now, this quadratic is clearly concave up so all that remains is to prove that it always lies ABOVE the x axis.

To do this, we let the discriminant be < 0 because it must have NO real roots in order to be ABOVE the x axis.

 

[asianese]

Note that in the case that the leading coefficient of the quadratic is unknown, the additional criteria that the leading coefficient is positive must also be satisfied.

 

[rumbleroar]

We want it to be an increasing function, so as you said we must have f'(x)>0.

We now have a quadratic in terms of K, and this quadratic is f'(x). We want this quadratic to be positive (meaning it is concave up and lies ABOVE the x axis). What I mean by a quadratic being positive is that all the Y values are positive.

Now, this quadratic is clearly concave up so all that remains is to prove that it always lies ABOVE the x axis.

To do this, we let the discriminant be < 0 because it must have NO real roots in order to be ABOVE the x axis.

Ahhh!! THANK YOU!! I see how they sneakily slipped in the positive/negative definites in this

Note that in the case that the leading coefficient of the quadratic is unknown, the additional criteria that the leading coefficient is positive must also be satisfied.

Thanks for the tip too!! conditions always trip me up