g(x)=sqrt (4x+1) and g(x)=5 a Solve the equation g(x)=5 b State the domain and range of the function.

a) To solve the equation $g(x)=5$, we need to find the value of $x$ for which $g(x)=5$.<br /><br />Given that $g(x)=\sqrt{4x+1}$, we can substitute this into the equation:<br /><br />$\sqrt{4x+1}=5$<br /><br />To solve for $x$, we can square both sides of the equation:<br /><br />$(\sqrt{4x+1})^2=5^2$<br /><br />$4x+1=25$<br /><br />$4x=24$<br /><br />$x=6$<br /><br />Therefore, the solution to the equation $g(x)=5$ is $x=6$.<br /><br />b) To state the domain and range of the function $g(x)=\sqrt{4x+1}$, we need to consider the restrictions on the input values (domain) and the possible output values (range).<br /><br />Domain:<br />The domain of the function $g(x)=\sqrt{4x+1}$ is the set of all real numbers $x$ for which the expression under the square root is non-negative. In other words, the domain is the set of all $x$ such that $4x+1 \geq 0$.<br /><br />Solving the inequality $4x+1 \geq 0$, we get:<br /><br />$4x \geq -1$<br /><br />$x \geq -\frac{1}{4}$<br /><br />Therefore, the domain of the function $g(x)=\sqrt{4x+1}$ is $x \geq -\frac{1}{4}$.<br /><br />Range:<br />The range of the function $g(x)=\sqrt{4x+1}$ is the set of all possible output values. Since the square root function always produces non-negative values, the range of the function is $y \geq 0$.<br /><br />In summary:<br />a) The solution to the equation $g(x)=5$ is $x=6$.<br />b) The domain of the function $g(x)=\sqrt{4x+1}$ is $x \geq -\frac{1}{4}$, and the range is $y \geq 0$.