How do you use the first and second derivitive to determine max and mins, inflection points, concavity, etc?

Question

I just don't get it. Please help me!

Answer 1

The first derivative is the slope of the line that is tangent to the graph. If there is a maximum or minimum, the tangent line at that point will be horizontal, so its slope is 0.

The second derivative is 0 when there is an inflection point. An inflection point is a place on the graph where it changes the direction of the concavity.

In the two cases above, the converse is not necessarily true. That is, just because the first derivative is 0 doesn't mean there has to be a max/min there.

If the second derivative is positive, the graph is concave upward which means like a regular "U". If it's negative, it's concave downward like an upside-down "U".

Answer 2

f(x) is concave down on the area the position f "(x) < 0 f(x) is concave down on the area the position f "(x) > 0 The inflection element for this function is at f " (x) = 0 maximums/minimum ensue at the same time as f ' (x) = 0 or on the sting of the function period f(x) = x * (x^2 + a million) = x^3 + x f ' (x) = 3x^2 + a million f "(x) = 6x **************************************... f "(x) = 0 6x = 0 x = 0 (C) The inflection element for this function is at x = 0 **************************************... f "(x) > 0 6x > 0 x > 0 (A) f(x) is concave up on the area of x from 0 to -5,4 **************************************... f "(x) < 0 6x < 0 x < 0 (B) f(x) is concave down on the area of x from 3e9748144467444e78c3c22db31e96683e9748144467444e78c3c22db31e9668 to 0 **************************************... f "(x) = 0 3x^2 + a million = 0 3x^2 = 3e9748144467444e78c3c22db31e96681 x^2 = 3e9748144467444e78c3c22db31e96681/3 x = undefined no optimal or minimum for this function if the period of x isn't said but3e9748144467444e78c3c22db31e9668 for the period [3e9748144467444e78c3c22db31e96683e9748144467444e78c3c22db31e96683e9748144467444e78c3c22db31e96683e9748144467444e78c3c22db31e9668] through searching on the function graph caricature(or through attempting the accessible x) f(3e9748144467444e78c3c22db31e96683e9748144467444e78c3c22db31e9668) = 3e9748144467444e78c3c22db31e9668130 f(3e9748144467444e78c3c22db31e96683e9748144467444e78c3c22db31e9668) = 3e9748144467444e78c3c22db31e966868 f(3e9748144467444e78c3c22db31e96683) = 3e9748144467444e78c3c22db31e966830 f(3e9748144467444e78c3c22db31e96682) = 3e9748144467444e78c3c22db31e966810 f(3e9748144467444e78c3c22db31e96681) = 3e9748144467444e78c3c22db31e96682 f(0) = 0 f(a million) = 2 f(2) = 10 f(3) = 30 f(-5,4) = sixty 8 we would want to be able to finish that (D) The minimum for this function occurs at x=3e9748144467444e78c3c22db31e96683e9748144467444e78c3c22db31e9668 f(3e9748144467444e78c3c22db31e96683e9748144467444e78c3c22db31e9668)=3e9748144467444e78c3c22db31e96683e9748144467444e78c3c22db31e9668(3e9748144467444e78c3c22db31e96683e9748144467444e78c3c22db31e9668^2 +a million) = 3e9748144467444e78c3c22db31e96683e9748144467444e78c3c22db31e9668 (26) = 3e9748144467444e78c3c22db31e9668130 (E) the optimal for this function occurs at x=-5,4 f(-5,4)= -5,4(-5,4^2+a million)=sixty 8