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Functions - Domain and Range; Composition

Ex 1 Find the domain and range for $f(x)=\sqrt{9-x^2}$.

Sol $f$ is defined for $9-x^2\ge0$ or $(3-x)(3+x)\ge0$, and solving this inequality gives $-3\le x\le 3$, so the domain is $[-3,3]$.

Since $y=\sqrt{9-x^2}$ gives $y^2=9-x^2$ or $x^2+y^2=9$, the graph of $f$ is the top half of the circle $x^2+y^2=9$; so the range of f is the set of all y-coordinates of points on this semicircle, which is the interval $[0,3]$.

Ex 2 If $f(x)=x^2+4x$ and $g(x)=3x-5$, find $(f\circ g)(x)$ and $(g \circ f)(x)$.

Sol $(f\circ g)(x)=f(g(x))=f(3x-5)=(3x-5)^2+4(3x-5)=9x^2-30x+25+12x-20=
9x^2-18x+5$, and $(g \circ f)(x)=g(f(x))=g(x^2+4x)=3(x^2+4x)-5=3x^2+12x-5$.

Pr 1 Find the domain and range for $f(x)= 4-x^2$.

Pr 2 Find the domain and range for $f(x)= 5+\sqrt{9-x}$.

Pr 3 Find the domain and range for $f(x)=\frac{5x}{x-8}$.

Pr 4 Find the domain for $f(x)=\frac{4x-20}{\sqrt{36-x^2}}$.

Pr 5 If $f(x)=x^2+4$ and $g(x)=\sqrt{2x-3}$, find $(f\circ g)(x)$ and $(g \circ f)(x)$.

Pr 6 If $h(x)=(x^2+6x-4)^5$, find functions f and g (different from h) such that $h=f \circ g$.

Pr 7 If $g(x)=2x-9$, find a function f such that $(g \circ f)(x)=x$.

Pr 8 Find the domain for $f(x)=x\sqrt{25-x^2}$.

Pr 9 Find the domain for $f(x)=\sqrt{\frac{x^2-5x}{x^2-9}}$.





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Lawrence Marx 2002-07-11