![f(x)=\sqrt[]{6x+16}](https://img.qammunity.org/2023/formulas/mathematics/college/10gk4kygrtbhzfcxxqcn5t8f816c8ae5k4.png)
First, we need to solve for f(x + h). To do this, we only have to replace x with x + h, like this.
![\begin{gathered} f(x+h)=\sqrt[]{6(x+h)+16} \\ f(x+h)=\sqrt[]{6x+6h+16} \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/v2co4di7813ijaiaicw54la174pqaqjbvt.png)
Now, we have a value for f(x) and f(x+h). Let's substitute this data to the equation being asked.
![(f(x+h)-f(x))/(h)=\frac{\sqrt[]{6x+6h+16}-\sqrt[]{6x+16}}{h}](https://img.qammunity.org/2023/formulas/mathematics/college/bat05wzhxyus3ixq2r1yw3r7qnyrnjqduq.png)
To simplify:
![\begin{gathered} (f(x+h)-f(x))/(h)=\frac{\sqrt[]{6x+6h+16}-\sqrt[]{6x+16}}{h} \\ (f(x+h)-f(x))/(h)=\frac{\sqrt[]{2(3x+3h+8)}-\sqrt[]{2(3x+8)}}{h} \\ (f(x+h)-f(x))/(h)=\frac{\sqrt[]{2}(\sqrt[]{3x+3h+8})-\sqrt[]{2}(\sqrt[]{3x+8})}{h} \\ (f(x+h)-f(x))/(h)=\frac{\sqrt[]{2}(\sqrt[]{3x+3h+8}-\sqrt[]{3x+8})}{h} \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/fedrtti1vabrih08s0iehgr5ac3bi425po.png)
The simplified equation is the last equation above.