How do you determine whether the function #f(x)= -6 sqrt (x)# is concave up or concave down and its intervals?

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Answer 1 · 2015-07-28T16:14:20

Use calculus (the sign of the second derivative) or algebra/precalculus graphing techniques.

Calculus

In general, to investigate concavity of the graph of function #f#, we investigate the sign of the second derivative.

#f(x) = -6x^(1/2)#

Note first that the doamin of #f# is #[0,oo)#

#f'(x) = -3x^(-1/2)#

#f''(x) = 3/2 x^(-3/2) = 3/(2sqrtx^3)#

#f''(x)# is positive for all real #x#, so it is positive for all #x# in the domain of #f#.

The graph of #f# is concave up on #(0,oo)#.

(Intervals of concavity are generally given as open intervals.)

Algebra/Precalculus

The graph of the square root function looks like this:

graph{y = sqrtx [-10, 10, -5, 5]}

That graph is concave down.

Multiplying by #-6# reflects the graph across the #x# asix and stretches it vertically by a factor of #6#:

graph{y = -6sqrtx [-6.41, 25.63, -13.2, 2.82]}

The graph is concave up.