Question

Let g be a function defined by g(x)=Integral from -1 to x of [(t^3-t^2-6t)/√t^2 +7]
On which of the following intervals is g decreasing?

A. x≤-2 and 0≤x≤3
B. x≤-2 and x≥3
C. -2≤x≤0 and x≥3
D. -2≤x≤3
E. x≤-1

I'm confused as to whether this function actually includes an integral or not.

Answer 1

Just to be clear, the g(x) function is this right?
\[\Large g(x) = \int_{-1}^{x}\frac{t^3-t^2-6t}{\sqrt{t^2+7}}dt\]

Answer 2

If so, then you use the second fundamental theorem of calculus

http://mathworld.wolfram.com/FundamentalTheoremsofCalculus.html

to go from this

\[\Large g(x) = \int_{-1}^{x}\frac{t^3-t^2-6t}{\sqrt{t^2+7}}dt\]

to this

\[\Large g \ '(x) = \frac{x^3-x^2-6x}{\sqrt{x^2+7}}\]


you'll use `g ' (x)` to determine where g(x) is increasing or decreasing

Answer 3

Huh

Answer 4

where are you stuck?

Answer 5

Un momento señor

Answer 6

So, to determine if it's decreasing, I would just have to see if the slope is negative, right?

Answer 7

which is equivalent to seeing where `g ' (x)` is negative

Answer 8

The denominator of `g ' (x) ` is never negative. So you just need to focus on the numerator

Answer 9

So THAT's the issue. I was looking at the function's slope, but not as a derivative function itself...

Answer 10

The derivative represents the slope of the tangent lines, so they are related ideas.

Answer 11

So it'd be A

Answer 12

I agree. The answer is A

Answer 13

Thanks mate

Answer 14

no problem