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 $f\left(x\right)=3-5x+4x^2$.

Short answer

The value of $f\left(a\right)=4a^2-5a+3,f\left(a+h\right)=4a^2+4h^2+8ah-5a-5h+3,\frac{f\left(a+h\right)-f\left(a\right)}{h}=4h+8a-5$.

Step-by-step solution

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

From the given function,

$f\left(x\right)=3-5x+4x^2...\left(1\right)$

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

role="math" localid="1644569704080" $$\begin{gathered} f\left(a\right)=3-5\left(a\right)+4{\left(a\right)}^2\mathrm{....}\left(\text{Substitute}x=a\text{in}\left(1\right)\right) \\ =4a^2-5a+3 \\ f\left(a+h\right)=3-5\left(a+h\right)+4{\left(a+h\right)}^2\mathrm{....}\left(\text{Substitute}x=a+h\text{in}\left(1\right)\right) \\ =3-5a-5h+4a^2+4h^2+8ah \\ =4a^2+4h^2+8ah-5a-5h+3 \end{gathered}$$

Step 2. Subtract fa from fa+h and divide by h where h≠0.

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

role="math" localid="1644569854366" $$\begin{gathered} \frac{f\left(a+h\right)-f\left(a\right)}{h}=\frac{\left(4a^2+4h^2+8ah-5a-5h+3\right)-\left(4a^2-5a+3\right)}{h} \\ =\frac{4h^2+8ah-5h}{h} \\ =4h+8a-5 \end{gathered}$$

Step 3. Description of steps.

The value of

$f\left(a\right)=4a^2-5a+3,f\left(a+h\right)=4a^2+4h^2+8ah-5a-5h+3,\frac{f\left(a+h\right)-f\left(a\right)}{h}=4h+8a-5$.