Inflection Points/Concavity

[R5232]

I encountered this problem on a exam review page. It is concerning concavity and inflection points.

e[sup:84d6kj7t]x[/sup:84d6kj7t](x+3)(x-2)
--------------------------
(x+1)

I understand what to do to find the inflection point and concavity of a polynomial function, but I haven't seen rational functions used in this manner
in lecture or homework, I'm not sure what angle to take to start this.

Any help is appreciated.

 

[fasteddie65]

f(x) = e^x(x + 3)(x - 2)/(x - 1)

This is a mess to differentiate. It involves both product rule and quotient rule.

Looking at the graph on my calculator, I see that x = 2 and x = -3 are zeros. You can also see that from the numerator.

There is no value for x = 1, from the denominator.

The derivative is negative if x < -3 and if x > 2. The derivative is zero when x = -4.

Sorry, but I just don't have time to do the actual derivative(s).

 

[Deleted member 4993]

I encountered this problem on a exam review page. It is concerning concavity and inflection points.

           e[sup]x[/sup](x+3)(x-2)
f(x) =  ------------------------------
                        (x+1)

I understand what to do to find the inflection point and concavity of a polynomial function, but I haven't seen rational functions used in this manner
in lecture or homework, I'm not sure what angle to take to start this.

Any help is appreciated.

u(x) = e[sup:3o3vvwlq]x[/sup:3o3vvwlq](x+3)(x-2)

u'(x) = e[sup:3o3vvwlq]x[/sup:3o3vvwlq](x+3)(x-2) + e[sup:3o3vvwlq]x[/sup:3o3vvwlq](x-2) + e[sup:3o3vvwlq]x[/sup:3o3vvwlq](x+3)= e[sup:3o3vvwlq]x[/sup:3o3vvwlq](x^2 + x - 6 + x - 2 + x +3) = e[sup:3o3vvwlq]x[/sup:3o3vvwlq](x^2 + 3x - 5)

f'(x) = [e[sup:3o3vvwlq]x[/sup:3o3vvwlq](x^2 + 3x - 5)(x+1) - e[sup:3o3vvwlq]x[/sup:3o3vvwlq](x+3)(x-2)]/(x+1)^2

Now continue....