Quick Calculus Question...?

Question

Can someone please help?

Is g(x) = -12.5(1.72)^x

1. concave up and increasing
2. concave up and decreasing
3. concave down and increasing
4. concave down and decreasing

I can't figure it out...

Answer 1

Concave down/Decreasing

Answer 2

To determine intervals of increase and decrease, we take the derivative and then make it 0.

g'(x) = (-12.5) (1.72)^x [ln(1.72)]

Since g'(x) is always negative, g(x) is always decreasing.

For the second derivative,

g''(x) = (-12.5) (ln(1.72)) (1.72)^x ln(1.72)

Notice that g''(x) is always going to be negative (since almost every term is positive, and this includes (1.72)^x). That means the only term negating is the -12.5, and it does so.

Therefore, g(x) is always concave down.

The answer is 4) Concave down and decreasing.

Answer 3

To answer your question you need to take first and second derivatives.

g(x) = -12.5(1.72)^x

g'(x) = -12.5[(1.72)^x](ln 1.72) = -12.5(ln 1.72)[(1.72)^x] < 0
for all x. So there are no critical points.

So the curve is decreasing everywhere.

g''(x) = -12.5(ln 1.72)²[(1.72)^x] < 0
for all x. So it is concave down everywhere.

The answer is

4. concave down and decreasing

Answer 4

the first derivative is slope (increasing or decreasing)
the second derivate is concavity (up or down)

Take the derivatives and evaluate in your domain.

Answer 5

i wont help with answers to cheat at homework just a tip use logarithm to get that x then derivate to look for maximums and minumums and use second derivate to get if its increaseing or decreasing.