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

Jul 29, 2018

The domain of the function is the interval $x \in \left[- 2 , 2\right]$.

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

We can also note that $f \left(x\right) \ge 0$ and $f \left(- 2\right) = f \left(2\right) = 0$.

Calculate now:

$\frac{\mathrm{df}}{\mathrm{dx}} = - \frac{x}{\sqrt{4 - {x}^{2}}}$

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

As:

${\lim}_{x \to - {2}^{+}} \frac{\mathrm{df}}{\mathrm{dx}} = + \infty$

${\lim}_{x \to {2}^{-}} \frac{\mathrm{df}}{\mathrm{dx}} = - \infty$

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

The only critical point where:

$\frac{\mathrm{df}}{\mathrm{dx}} = 0$

is $x = 0$.

We can easily see that $\frac{\mathrm{df}}{\mathrm{dx}} > 0$ for $x < 0$ and $\frac{\mathrm{df}}{\mathrm{dx}} < 0$ 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 $f \left(0\right) = 2$ and it is clearly also the absolute maximum.

$\frac{{d}^{2} f}{\mathrm{dx}} ^ 2 = \frac{- \sqrt{4 - {x}^{2}} - {x}^{2} / \sqrt{4 - {x}^{2}}}{{x}^{2} - 4}$

$\frac{{d}^{2} f}{\mathrm{dx}} ^ 2 = \frac{- 4 + {x}^{2} - {x}^{2}}{\left({x}^{2} - 4\right) \sqrt{4 - {x}^{2}}}$

$\frac{{d}^{2} f}{\mathrm{dx}} ^ 2 = - \frac{4}{\left({x}^{2} - 4\right) \sqrt{4 - {x}^{2}}}$

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.