How do you find intercepts, extrema, points of inflections, asymptotes and graph y=xsqrt(16-x^2)y=x16x2?

2 Answers
Jan 11, 2017

f(x) = x sqrt(16-x^2)f(x)=x16x2 is defined for x in [-4,4]x[4,4], has no asymptotes, intercepts y=0y=0 at x_1=-4x1=4, x_2 = 0x2=0, x_3 = 4x3=4, has an absolute minimum for x_m=-2sqrt(2)xm=22 and absolute maximum at x_M= 2sqrt(2)xM=22 and an inflection point for x=0x=0.

Explanation:

Consider:

f(x) = xsqrt(16-x^2)f(x)=x16x2

The function is defined for 16-x^2 >=016x20 that is for x in [-4,4]x[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) = 0x16x2=0

which yields three roots:

x_1=-4x1=4, x_2 = 0x2=0, x_3 = 4x3=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 8xx16 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]}