Derivates and the Shape of Curves?

Question

(a) Find the intervals of increase or decrease
(b) Find the local max and min values
(c) Find the intervals of concavity and the inflection points.
(d) How would you sketch the graph using information from (a-c)

A(x)= xsqrt x + 3

Answer 1

(a) Derive your equation and find your critical points. In the intervals between those points, look at the slope and you'll be able to find out where your fuction increases and decreases.

(b) Again, look at your critical points and their coordinates and whether the intervals on either side of it are decreasing and increasing/increasing and decreasing.

(c) Look at how the slope changes on your intervals. If the interval is increasing and the slope is increasing, you have upwards concavity. But if the slope is decreasing, you have downwards concavity. For a decreasing interval, is the slope is increasing (look at the slope's absolute value), then you have downwards concavity; if it is decreasing (again, the abs. value), then you have upwards concavity. For inflections points, look for critical points that have intervals on either side that are either both increasing or decreasing.

(d) Find the coordinates of your critical points, and use concavity and such to see how to "connect the dots."

Hope this helped (I'm not doing your homework all the way for you.) Good luck!

Hint: check the graph on a graphing calculator if you get stuck, but don't try to use it as a crutch. Remember, a calculator is a tool, not an "easy button", so don't try to treat it otherwise or you'll regret it later.

Answer 2

The function sqrt(x) is real for x greater than or equal to zero. For values of x less than zero, the sqrt is imaginary.

the function will increase for all values of x greater than zero

The min is x=0

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