How do you find intercepts, extrema, points of inflections, asymptotes and graph #y=xsqrt(16-x^2)#?

2 Answers
Jan 11, 2017

#f(x) = x sqrt(16-x^2)# is defined for #x in [-4,4]#, has no asymptotes, intercepts #y=0# at #x_1=-4#, #x_2 = 0#, #x_3 = 4#, has an absolute minimum for #x_m=-2sqrt(2)# and absolute maximum at #x_M= 2sqrt(2)# and an inflection point for #x=0#.

Explanation:

Consider:

#f(x) = xsqrt(16-x^2)#

The function is defined for #16-x^2 >=0# that is for #x in [-4,4]# and it is continuous in its domain.

At the function is continuous in a compact domain, it cannot have asymptotes.

(1) We can find the intercepts solving the equation:

#xsqrt(16-x^2) = 0#

which yields three roots:

#x_1=-4#, #x_2 = 0#, #x_3 = 4#

(2) Calculate the first derivative:

#f'(x) = sqrt(16-x^2) -x^2/sqrt(16-x^2)= (16-x^2-x^2)/sqrt(16-x^2) = (16-2x^2)/sqrt(16-x^2)#

so #f(x)# is only differentiable in the open interval #x in (-4,4)#, and the critical points can be determined as:

#f'(x) = 0#

#(16-2x^2)/sqrt(16-x^2)=0#

#16-2x^2 = 0#

#x= +-2sqrt(2)#

Solving the inequality:

#f'(x) > 0#

we can see that:

#(16-2x^2)/sqrt(16-x^2)>0# for #abs (x) < 2sqrt(2)#

This means that #f(x)# is:

  • decreasing in the interval #x in [-4,-2sqrt(2)]#

  • increasing in the interval #x in [-2sqrt(2),2sqrt(2)]#

  • decreasing in the interval #x in [2sqrt(2),4]#

and thus:

  • #x_m = -2sqrt(2)# is a local minimum
  • #x_M = 2sqrt(2)# is a local maximum

Based on Weierstrass' Theorem, as #f(x)# is continuous in a compact domain it also has absolute minimum and maximum, and we can easily check that #x_m# and #x_M# are the absolute minimum and maximum respectively.

(3) We calculate the second derivative:

#f''(x) = frac (-4x sqrt(16-x^2) +x (16-2x^2) /sqrt(16-x^2)) (16-x^2)= frac (-4x (16-x^2) +x (16-2x^2) ) ((16-x^2)^(3/2))= frac (x (-64+4x^2+16-2x^2) ) ((16-x^2)^(3/2)) = frac (2x (x^2-24) ) ((16-x^2)^(3/2))#

and we can see that #f(x)#has only one inflection point in its domain at #x=0#, since #x^2-24 < 0# for #x in (-4,4)#, so that:

  • #f''(x) > 0# for #x in (-4,0)# and so #f(x)# is concave up.
  • #f''(x) < 0# for #x in (0,4)# and so #f(x)# is concave down.

graph{x sqrt(16-x^2) [-20, 20, -10, 10]}

Jan 11, 2017

See explanation.

Explanation:

To make y real, #x in [-4, 4]#

x-intercept ( y = 0 ) : #+-4#

The graph also passes through the origin

Upon substitution

#x = 4 sin theta#,

#y = 16 sin theta cos theta = 8 sin 2theta#, revealing that

#y in [-8, 8].#.

The graph is contained in an 8#xx#16 rectangle, and so, the

question of search for asymptote(s) does not arise.

#y'=((dy)/(d theta))/((dx)/(d theta))=(16 cos 2theta)/(4 cos theta)#

#=4cos2theta sec theta#

=0, at #theta=+-pi/4 to x = +-2sqrt 2=+-2.83#, nearly.

Max y = 8, at #theta = pi/4#

Min y = #- 8#, at #theta=-pi/4#.

#y''=((dy')/(d theta))/((dx)/(d theta)#

#=4( (-2sin 2theta sec theta+cos 2theta sec theta tan theta))#

#/(4cos theta)#

=0, only when #theta = 0# ( after simplification, sin theta is a factor in

the numerator ).

The origin is the point of inflexion ( after studying tangent-crossing -

curve, at #theta = 0#.

graph{(y-xsqrt(16-x^2))(y-8.1)(y+8.1)=0[-5, 5, -9, 9]}