# How do you use the first and second derivatives to sketch  y = x + (1-x)^(1/2)?

Aug 9, 2017

The derivatives of a function can help with plotting them because they indicate inflection points and direction of the curve.

#### Explanation:

If the first derivative of the equation is positive at that point, then the function is increasing. If it is negative, the function is decreasing.
http://mathworld.wolfram.com/FirstDerivativeTest.html

If f(x,y) is a two-dimensional function that has a local extremum at a point (x_0,y_0) and has continuous partial derivatives at this point, then f_x(x_0,y_0)=0 and f_y(x_0,y_0)=0. The second partial derivatives test classifies the point as a local maximum or local minimum.
Weisstein, Eric W. "Second Derivative Test." From MathWorld--A Wolfram Web Resource. http://mathworld.wolfram.com/SecondDerivativeTest.html

Aug 10, 2017

The first step I would do when sketching a curve is to find intercepts. These will occur when x is $0$ (y-intercepts) and when $y$ is $0$ (x-intercepts).

${y}_{\text{intercept}} = 0 + {\left(1 - 0\right)}^{\frac{1}{2}} = \pm 1$

The x-intercepts can be a little bit more complicated to find.

$0 = x + {\left(1 - x\right)}^{\frac{1}{2}}$

$- x = {\left(1 - x\right)}^{\frac{1}{2}}$

If we square both sides, we will be able to get rid of the √.

${\left(- x\right)}^{2} = {\left({\left(1 - x\right)}^{\frac{1}{2}}\right)}^{2}$

${x}^{2} = 1 - x$

${x}^{2} + x - 1 = 0$

$x = \frac{- 1 \pm \sqrt{{1}^{2} - 4 \cdot 1 \cdot - 1}}{2 \cdot 1}$

$x = \frac{- 1 \pm \sqrt{5}}{2}$

However, $\frac{- 1 + \sqrt{5}}{2}$ will be extraneous because the function isn't defined for any $x$ above $1$.

The next step in curve sketching is to find the first derivative. This will give us insight into the direction in which the graph is going and where maximums and minimums will occur. Obtained using the chain rule, we get:

$y ' = 1 - \frac{1}{2 \sqrt{1 - x}}$

Critical points occur whenever the derivative equals $0$ or is undefined (obviously if a critical point occurs when the function is undefined, it is not a critical point).

The derivative is undefined at $x = 1$ (but the function is defined there. We solve now for when the derivative equals $0$.

$0 = 1 - \frac{1}{2 \sqrt{1 - x}}$

$\frac{1}{2 \sqrt{1 - x}} = 1$

$1 = 2 \sqrt{1 - x}$

$1 = 4 \left(1 - x\right)$

$1 = 4 - 4 x$

$- 3 = - 4 x$

$\frac{3}{4} = x$

This does indeed work in the original equation. Next, we must verify the intervals of increase, decrease, by selecting test points in the intervals $\left(- \infty , \frac{3}{4}\right) , \left(\frac{3}{4} , 1\right) \mathmr{and} \left(1 , \infty\right)$.

test point 1: $x = 0$

$y ' \left(0\right) = 1 - \frac{1}{2 \sqrt{1 - 0}} = 1 - \frac{1}{2} = \frac{1}{2}$

Since this is positive, the function is increasing on $\left(- \infty , \frac{3}{4}\right)$, and $x = \frac{3}{4}$ is hence a maximum. It has to be determined whether it's an absolute or a local max.

test point 2: $x = \frac{4}{5}$

$y ' \left(\frac{4}{5}\right) = 1 - \frac{1}{2 \sqrt{1 - \frac{4}{5}}} = \text{negative}$

I don't even bother computing here, because all that matters is the sign.

Hence, $x = 1$ will be a minimum (we don't know if it will be a local max or an absolute max just yet).

Next, we examine end behaviour. We notice that the domain of this function is {x| x ≤ 1, x in RR}. If we select values of $x$ that are becoming bigger and bigger in the negative direction, we notice that $y$ is tending to $- \infty$, therefore, $x = 1$ is a local minimum and $x = \frac{3}{4}$ is an absolute maximum.

We finally must determine the second derivative in order to look at concavity and points of inflection. The second derivative is given by differentiating the first. An alternative writing of the first derivative is

$y ' = 1 - \frac{1}{2} {\left(1 - x\right)}^{- \frac{1}{2}}$

This can be differentiated using the chain rule.

$y ' ' = 0 - \frac{1}{2} \left(- \frac{1}{2}\right) \left(- 1\right) {\left(1 - x\right)}^{- \frac{3}{2}}$

y'' = -1/(4(1 - x)^(-3/2)

We see immediately if we set $y ' '$ to $0$, there is no solution for $x$ to the resulting equation. So there are no inflection points, that's to say the sign of the second derivative remains the same on the entire domain of the function.

If we select a few test points, we notice that all the outputs on $y ' '$ are negative, which means the graph is concave down (convex), on {x|x ≤ 1, x in RR}. So, after all this work we can trace the graph quite precisely, as I have done on the following grid. Hopefully this helps!