Solution.

We note that \(x=5\), we should be dividing by zero which is not defined since infinity \((\infty )\) is not a real number.
To come up with the domain we must remove that which causes problem.

\(D_f=(-\infty ,5)\cup (5,\infty )\) or \(D_f=\{x\in \mathbb {R}:x\neq 5\}\) or \(D_f=\mathbb {R}-\{5\}\).

Solution.

1.

Here every real number will produce another real number. Therefore there is no real number which causes problems.
Hence \(D_f=\mathbb {R}\) the set of all real numbers.

2.

So we note that if we replace \(x\) by a negative number then we get an imaginary number which is not a real number. Therefore to find the domain we must remove the set of any negative real numbers.

\(D_g=[0,\infty )\)

\(D_g=\{x\in \mathbb {R}:x\geq 0\}\)

3.

Here \(x^{2}-4=0\) at \(x=2\) and \(x=-2\), so both must be removed from \(\mathbb {R}\).

\(D_h=(-\infty ,-2)\cup (-2,2)\cup (2,\infty )\)

\(D_h=\mathbb {R}-\{-2,2\}\)

Solution.

\(f(-4)=(-4)^2+1=16+1=17=f(4)\)

\(f(-3)=(-3)^2+1=9+1=10=f(3)\)

\(f(-2)=(-2)^2+1=4+1=5=f(2)\)

\(f(-1)=(-1)^2+1=1+1=2=f(1)\)

\(f(0)=0^2+1=0+1=1\)

Images are 1, 2, 5, 10, 17 \(\implies R_f=\{1,2,5,10,17\}\)

Solution. For the domain ask what would break; for the range ask what values the formula can actually produce.

(i). \(f(x)=3x+2\). Nothing breaks, so \(D_f=\mathbb {R}\). Any real \(y\) is reached, by \(x=\frac {y-2}{3}\), so \(R_f=\mathbb {R}\).

(ii). \(g(x)=2+\sqrt {2x-3}\). The root needs \(2x-3\geq 0\), so \(D_g=\left [\frac {3}{2},\infty \right )\). A square root is never negative and takes every value from \(0\) upwards, so \(g\) takes every value from \(2\) upwards: \(R_g=[2,\infty )\).

(iii). \(h(x)=\frac {2}{x^{2}+1}\). Since \(x^{2}+1\geq 1>0\) for every real \(x\), nothing is excluded and \(D_h=\mathbb {R}\). As \(x^{2}+1\) ranges over \([1,\infty )\), its reciprocal ranges over \((0,1]\), so \(h\) ranges over \((0,2]\). The value \(2\) is attained at \(x=0\); the value \(0\) is approached but never reached, which is why the bracket there is round: \(R_h=(0,2]\).

(iv). \(p(x)=\left |x-2\right |-3\). Nothing breaks, so \(D_p=\mathbb {R}\). The modulus is at least \(0\) and takes every non-negative value, so \(p\) takes every value from \(-3\) upwards, the minimum occurring at \(x=2\): \(R_p=[-3,\infty )\).