How do you find intercepts, extrema, points of inflections, asymptotes and graph #h(x)=xsqrt(9-x^2)#?

1 Answer
Aug 22, 2017

Please see below.

Explanation:

Domain : #[-3,3]#

Intercepts
#y# intercept #h(0) = 0#

#x# intercepts: #-3#, #0#, and #3#
(Solve #h(x) = 0#)

Asymptotes
The function has no horizontal asymptotes. (The domain is bouinded.) and no vertical asymptotes (there is no division).

#h'(x) = (9-2x^2)/sqrt(9-x^2)#

Is undefined only at the endpoints of the domain and is #0# at #x = +-3/sqrt2#

It is worth noting that as #x# approaches the endpoints of the domain, #+-3#, the derivative goes to #+-oo# So the tangent line becomes vertical.

Local extrema and increasing/decreasing
On #[-3,-3/sqrt2)#, we have #h'(x) < 0# so #h# is decreasing.
On #(-3/sqrt2,0)#, we have #h'(x) > 0# so #h# is increasing.

And #h(-3/sqrt2) = -9/2# is a local minimum.

On #(0,3/sqrt2)#, we have #h'(x) > 0# so #h# is increasing.
On #(3/sqrt2,3]#, we have #h'(x) ><0# so #h# is decreasing.

And #h(3/sqrt2) = 9/2# is a local maximum.

Concavity and inflection points

#h''(x) = (x(2x^2-27))/(9-x^2)^(3/2)#

Is undefined only at the endpoints of the domain and is #0# at #x = 0# (The expression is also #0# at #x = +-sqrt(27/2)#, but those are outside the domain of #h#.)

On #[-3,0)#, we have #h''(x) > 0# so the graph of #h# is concave upwards (convex).

On #[-3,0)#, we have #h''(x) < 0# so the graph of #h# is concave downwards (concave).

The point #(0,0)# is the only inflection point.

The graph is:

graph{xsqrt(9-x^2) [-9.23, 13.27, -5.625, 5.625]}