Question

7. f(x) = x² + 4 (a) f(-2) (b) f(3) f) f (c) f(2) (d) f(x + bx)

Answer 1

`\begin{gathered} a)\text{ 2}\sqrt[]{2} \\ b)\text{ }\sqrt[]{13} \\ c)\text{ 2}\sqrt[]{2} \\ d)\text{ }\sqrt[]{x^2(1+2b+b^2)+4}\text{ } \end{gathered}`Step-by-step explanation:
`f(x)\text{ = }\sqrt[]{x^2\text{ + 4}}`
`\begin{gathered} a)\text{ when x = -2} \\ f(-2)\text{ = }\sqrt[]{(-2)^2\text{ + 4}}\text{ = }\sqrt[]{4+4} \\ f(-2)\text{ = }\sqrt[]{8}\text{ = }\sqrt[]{2*4} \\ f(-2)\text{ = 2}\sqrt[]{2} \end{gathered}`
`\begin{gathered} b)\text{ }when\text{ x = 3} \\ f(3)\text{ = }\sqrt[]{(3)^2+\text{ 4}}\text{ = }\sqrt[]{9\text{ + 4}} \\ f(3)\text{ = }\sqrt[]{13} \end{gathered}`
`\begin{gathered} c)\text{ when x = 2} \\ f(2)\text{ = }\sqrt[]{(2)^2+4}\text{ = }\sqrt[]{4+4} \\ f(2)\text{ = }\sqrt[]{8} \\ f(2)\text{ = 2}\sqrt[]{2} \end{gathered}`
`\begin{gathered} d)\text{ when x = x + bx} \\ f(x\text{ + bx) = }\sqrt[]{(x+bx)^2+4}\text{ = }\sqrt[]{(x+bx)(x^{}+bx)+4} \\ f(x\text{ + bx) = }\sqrt[]{(x^2+bx^2+bx^2+b^2x^2)+4} \\ f(x\text{ + bx) = }\sqrt[]{(x^2+2bx^2^{}+b^2x^2)+4} \\ f(x\text{ + bx) = }\sqrt[]{x^2(1+2b+b^2)+4}\text{ } \end{gathered}`