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9 Functions of the Form \(f(x)=\sqrt {x}\)

We have already seen that the function \(f(x)=\sqrt {x}\) is defined only for values of \(x\geq 0\). Only non-negative values will appear for \(x\).

\(x\) 0 1 4 9 16
\(f(x)=\sqrt {x}\) 0 1 2 3 4

1.
Thus all the graphs of the form \(f(x)=\sqrt {x}\) have the same shape of the graph below
√ -- xf ≥(x0) = x
2.
\(f(x)=-\sqrt {x}\). This is a reflection of the graph of the function \(f(x)=\sqrt {x}\) in the
\(x-\) axis.
√ -- xf(≥x)0= − x
3.
\(f(x)=\sqrt {-x}\). The domain of this function is the set of all non positive real numbers \(x\leq 0\), its graph below
√ --- xf ≤(x)0 = − x
4.
\(f(x)=-\sqrt {-x}\). This function is a reflection of \(f(x)=\sqrt {-x}\) in the \(x-\) axis.
√--- xf(≤x)0= − − x

Consider now the graph of \(f(x)=\sqrt {x-2}\). To find the domain of \(f\) we must solve \[x-2\geq 0 \implies x\geq 2 ,\] so \(D_f=[2,\infty )\).

2

The range of \(f\) is all the non-negative real numbers.

The graph of \(f(x)=\sqrt {x-2}\) has the same shape as the graph of \(f(x)=\sqrt {x}\). Except that the graph will start at \(x=2\).

√ ----- |2f (x ) = x − 2

Example 9.1.

Sketch \(f(x)=-1+\sqrt {x}\).

Solution.

The domain \(D_f\) is \(x\geq 0\).

The range \(R_f\) is \(-1\) to \(\infty \). i.e \(R_f=[-1,\infty )\)

√ -- |1−−f 1(x ) = − 1+ x

Example 9.2.

State the domain and the range of each of the functions below and hence sketch the graph of the function.

1.
\(f(x)=-\sqrt {2x-1}\)

Domain:
\(2x-1\geq 0\implies x\geq \frac {1}{2}\).

\(\therefore D_f=\left [\frac {1}{2},\infty \right )\)

Since the sign in front of the root symbol is negative the range of \(f\) must be negative numbers.
But when \(x=\frac {1}{2}, y=0\).
Therefore the starting point is \(\left (\frac {1}{2},0\right ).\)

√ ------ |12f(x) = − 2x− 1
2.
\(g(x)=-1+\sqrt {3-x}\)

Domain of \(g\): \(3-x\geq 0\implies 3\geq x\)

\(D_g=(-\infty ,3]\)

Most of the values in the range will be positive.
However, when \(x=3, y=-1\)
The graph starts at \((3,-1)\).
We note that the graph will cross both \(y=-1+\sqrt {3-x}\) but where it crosses the \(x-\)axis, \(y=0\) thus \(0=-1+\sqrt {3-x}\implies 1=\sqrt {3-x}\implies 3-x=1\implies x=2\).

It crosses the \(x-\)axis at \((2,0)\) and it crosses the \(y-\)axis when \(x=0, f(0)=-1+\sqrt {3}\).
It crosses the \(y-\)axis at \((0,-1+\sqrt {3})\)

√ -- √ ----- |23(f−−−3(,x11−)+ 1=)−3 1+ 3− x
3.
\(P(x)=2-\sqrt {1+2x}\)

\(D_f=\left [-\frac {1}{2},\infty \right )\)

starts at \(\left (-\frac {1}{2},2\right )\)

\(x-\)intercept, \(y=0\)
\(0=2-\sqrt {1+2x}\implies 1+2x=4\implies 2x=4-1\implies 2x=3\implies x=\frac {3}{2}\)

crosses the \(x-\)axis at \(\left (\frac {3}{2},0\right )\)

\(y-\)intercept \(x=0\)
\(P(0)=2-\sqrt {1+0}=2-1=1\)

crosses \(y-\)axis at \((0,1)\).

3 1 √ ------ |2f−1−2(x2) = 2− 1 + 2x
4.
\(h(x)=3\sqrt {3x+2}+2\)

Domain: \(3x+2\geq 0\implies x\geq -\frac {2}{3}\), so \(D_h=\left [-\frac {2}{3},\infty \right )\).

Range: \(\sqrt {3x+2}\geq 0\), so \(3\sqrt {3x+2}\geq 0\) and \(h(x)\geq 2\). Hence \(R_h=[2,\infty )\).

starts at \(\left (-\frac {2}{3},2\right )\)

\(x-\)intercept, \(y=0\)
\(0=3\sqrt {3x+2}+2\implies 3\sqrt {3x+2}=-2\), which is impossible because a square root is never negative. There is no \(x-\)intercept — consistent with the range, since \(h\) never drops below \(2\).

\(y-\)intercept, \(x=0\)
\(h(0)=3\sqrt {2}+2\approx 6.24\)

crosses the \(y-\)axis at \(\left (0,3\sqrt {2}+2\right )\).

√ -- √------ −−2−3h(23x2)+=23 3x + 2 + 2

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