Question

Find the value of $f\left(a\right),f\left(a+h\right),\frac{f\left(a+h\right)-f\left(a\right)}{h}$for the function $f\left(x\right)=3x^2+2$.

Short answer

The value of $f\left(a\right)=3a^2+2,f\left(a+h\right)=3a^2+6ah+3h^2+2,\frac{f\left(a+h\right)-f\left(a\right)}{h}=6a+3h$.

Step-by-step solution

Step 1. Evaluate the values of fa and fa+h.

From the given function,

$f\left(x\right)=3x^2+2....\left(1\right)$

Evaluate $f\left(a\right)\text{and}f\left(a+h\right)$gives,

role="math" localid="1644563800473" $$\begin{gathered} f\left(a\right)=3{\left(a\right)}^2+2....\left(\text{Substitute}x=a\text{in}\left(1\right)\right) \\ =3a^2+2 \\ f\left(a+h\right)=3{\left(a+h\right)}^2+2....\left(\text{Substitute}x=a+h\text{in}\left(1\right)\right) \\ =3a^2+6ah+3h^2+2 \end{gathered}$$

Step 2. Subtract fa from fa+h and divide by h.

Evaluate $\frac{f\left(a+h\right)-f\left(a\right)}{h}$gives,

role="math" localid="1644563960950" $$\begin{gathered} \frac{f\left(a+h\right)-f\left(a\right)}{h}=\frac{\left(3a^2+6ah+3h^2+2\right)-\left(3a^2+2\right)}{h} \\ =\frac{3a^2+6ah+3h^2+2-3a^2-2}{h} \\ =6a+3h \end{gathered}$$

Step 3. Description of steps.

The value of $f\left(a\right)=3a^2+2,f\left(a+h\right)=3a^2+6ah+3h^2+2,\frac{f\left(a+h\right)-f\left(a\right)}{h}=6a+3h$.