Question

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

Answer 1

`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]}

Answer 2

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]}