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

##### 2 Answers

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.

See Also:

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

The first step I would do when sketching a curve is to find intercepts. These will occur when x is

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

#0 = x + (1 - x)^(1/2)#

#-x = (1 - x)^(1/2)#

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

#(-x)^2 = ((1 - x)^(1/2))^2#

#x^2 = 1 - x#

#x^2 + x - 1 = 0#

By the quadratic formula:

#x = (-1 +- sqrt(1^2 - 4 * 1 * -1))/(2 * 1)#

#x= (-1 +- sqrt(5))/2#

However,

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 -1/(2sqrt(1 - x))#

Critical points occur whenever the derivative equals

The derivative is undefined at

#0 = 1- 1/(2sqrt(1 - x))#

#1/(2sqrt(1 - x)) = 1#

#1 = 2sqrt(1 - x)#

#1 = 4(1 - x)#

#1 = 4 - 4x#

#-3 = -4x#

#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

**test point 1: #x = 0#**

#y'(0) = 1 - 1/(2sqrt(1 - 0)) = 1 - 1/2 = 1/2#

Since this is positive, the function is increasing on

**test point 2: #x = 4/5#**

#y'(4/5) = 1 - 1/(2sqrt(1 - 4/5))= "negative"#

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

Hence,

Next, we examine end behaviour. We notice that the domain of this function is

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 - 1/2(1 - x)^(-1/2)#

This can be differentiated using the chain rule.

#y'' = 0 - 1/2(-1/2)(-1)(1 - x)^(-3/2)#

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

We see immediately if we set

If we select a few test points, we notice that all the outputs on

Hopefully this helps!