Decreasing Function

[mathshelpplease]

I am doing this problem and am a little stuck on part b.


I have worked out the f'(x) which is equal to

having double checked through derivative-calculator.net

and as it I need the interval with strictly decreasing, have determined [math]f'(x)≤ 0[/math]
But am struggling to work out how to determine the values for such a complicated function.
I know that exp will be positive, but other than that any advice appreciated!

 

[Steven G]

You have 3 factors in the numerator. The first factor is 2sqrt(3) which is always positive. The 2nd factor is a quadratic equation. The third factor is an exponential function which is always positive. The denominator is also non-negative.

It should be clear that the final sign will depend on the 2nd factor.
Why are you having trouble determine when a quadratic equation is negative?

 

[mathshelpplease]

You have 3 factors in the numerator. The first factor is 2sqrt(3) which is always positive. The 2nd factor is a quadratic equation. The third factor is an exponential function which is always positive. The denominator is also non-negative.

It should be clear that the final sign will depend on the 2nd factor.
Why are you having trouble determine when a quadratic equation is negative?

Thank you this makes a lot of sense. I am new to this area so think that is why I am finding it particularly difficult.

Following from what you've said, I factorised the second factor to get [math](sqrt(3)+1)(x-2)(x-1)[/math]
Meaning that the critical values of x will be 2 and 1, as the first factor will always be positive.

I then did a sign diagram which followed

	x<1	x=1	1<x<2	x=2	2<x
x-2	-	-	-	0	+
x-1	-	0	+	+	+
f'(x)	+	0	-	0	+
f(x)	increasing		decreasing		increasing

Does this mean that the overall answer will be decreasing through 1<x<2 ?

I was wondering if I actually needed to differentiate further, or have made an error in my method.

Thank you

 

[Steven G]

The table you did is fine and f(x) is decreasing for 1<x<2.