How do you use the first and second derivatives to sketch f(x) = sqrt(4 - x^2)?

1 Answer
Jul 29, 2018

The domain of the function is the interval x in [-2,2].

We can note that the function is even, that is f(-x) = f(x) so its graph will be symmetrical with respect to the y axis.

We can also note that f(x) >=0 and f(-2) = f(2) = 0.

Calculate now:

(df)/dx = -x/sqrt(4-x^2)

which means that the function is differentiable only in the interior of the interval.

As:

lim_(x->-2^+) (df)/dx = +oo

lim_(x->2^-) (df)/dx = -oo

the tangent to the graph at the limits of the interval of definition is vertical.

The only critical point where:

(df)/dx = 0

is x=0.

We can easily see that (df)/dx >0 for x <0 and (df)/dx < 0 for x > 0, so the function will be strictly increasing from x=-2 to x=0 where it has a local maximum and strictly decreasing from x=0 to x=2.

The value of the local maximum is f(0) = 2 and it is clearly also the absolute maximum.

(d^2f)/dx^2 = (-sqrt(4-x^2)-x^2/sqrt(4-x^2))/(x^2-4)

(d^2f)/dx^2 = (-4+x^2-x^2)/((x^2-4)sqrt(4-x^2))

(d^2f)/dx^2 = -4/((x^2-4)sqrt(4-x^2))

The second derivative is always negative, so the function is concave down in its entire domain.

graph{sqrt(4-x^2) [-4.933, 4.932, -2.466, 2.467]}

If you let:

y = sqrt(4-x^2)

then:

y^2 = 4-x^2

x^2+y^2 = 4

so the graph is the part of the circle of center in the origin and radius r = 2 above the x axis.