# HOW TO SKETCH SQUARE ROOT FUNCTION

When we are sketching the graph of functions either by hand or through any graphing software we cannot show the entire graph. Only a part of the graph can be sketched.

A crucial question is which part of the curve we need to show and how to decide that part. To decide on this we use the derivatives of functions. We enlist few guidelines for determining a good viewing rectangle for the graph of a function. They are :

(i) The domain and the range of the function.

(ii) The intercepts of the cure (if any).

(iii) Critical points of the function.

(iv) Local extrema of the function.

(v) Intervals of concavity.

(vi) Points of inflections (if any).

(vii) Asymptotes of the curve (if exists)

Sketch the graphs of the following functions.

Problem 1 :

y  =  x√(4-x)

Solution :

Domain and range :

Possible values of x are (-∞, 3) ==>  Domain

y values are (-∞, 4) ==>  Range

Intercepts :

To find x intercept, put y  =  0

x√(4-x)  =  0

 x  =  0 √(4-x)  =  04-x  =  0x  =  4

x intercepts are 0 and 4.

y-intercept :

Put x  =  0

y  =  0

Critical points :

f(x)  =  x√(4-x)

f'(x)  =  x(1/2√(4-x))(-1) + √(4-x) (1)

f'(x)  =  -x/2√(4-x) + √(4-x)

f'(x)  =  [-x+2(4-x)]/2√(4-x)

f'(x)  =  (-x+8-2x)/2√(4-x)

f'(x)  =  (8-3x)/2√(4-x)

f'(x)  =  0

(8-3x)/2√(4-x)  =  0

3x  =  8

x  =  8/3

Plotting these points on the number line and dividing into intervals, we get (-∞, 8/3) and (8/3, ∞).

Local extrema :

f'(x)  =  (8-3x)/2√(4-x)

u  =  8-3x and v  =  2√(4-x)

u'  =  -3 and v'  =  -1/√(4-x)

f''(x)  =  [2√(4-x)(-3)-(8-3x)(-1/√(4-x))] /4(4-x)

f''(x)  =  (3x-16)/4(4-x)√(4-x)

f''(8/3)  =  - < 0 (Maximum)

Maximum value at x  =  8/3 :

y  =  x√(4-x)

y  =  (8/3)√(4-(8/3))

y  =  (8/3)√(4/3)

y  =  16/33

So, the maximum point is (8/3, 16/33).

Intervals of concavity :

f''(x)  =  (3x-16)/4(4-x)√(4-x)

f''(x)  =  0

3x-16  =  0

x  =  16/3

It is not in the domain, so it has no real value on.

Point of inflection :

There is no point of inflection.

Asymptotes :

As x ->∞, y -> ±

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