use the formula $f(x)=limlimits_{z o x}\frac{f(z)-f(x)}{z - x}$ to find the derivative of the following function.

$f(x)=2x^{2}-5x + 5$

$f(x)=\square$

Question

use the formula $f(x)=limlimits_{z	o x}\frac{f(z)-f(x)}{z - x}$ to find the derivative of the following function.

$f(x)=2x^{2}-5x + 5$

$f(x)=\square$

Accepted Answer

# Explanation:
## Step1: Substitute $f(z)$ and $f(x)$
Given $f(x)=2x^{2}-5x + 5$, then $f(z)=2z^{2}-5z + 5$.
$$
\begin{align*}
\frac{f(z)-f(x)}{z - x}&=\frac{(2z^{2}-5z + 5)-(2x^{2}-5x + 5)}{z - x}\
&=\frac{2z^{2}-5z+5 - 2x^{2}+5x - 5}{z - x}\
&=\frac{2(z^{2}-x^{2})-5(z - x)}{z - x}
\end{align*}
$$
## Step2: Factor and simplify
Use the formula $a^{2}-b^{2}=(a + b)(a - b)$, so $z^{2}-x^{2}=(z + x)(z - x)$.
$$
\begin{align*}
\frac{2(z^{2}-x^{2})-5(z - x)}{z - x}&=\frac{2(z + x)(z - x)-5(z - x)}{z - x}\
&=\frac{(z - x)[2(z + x)-5]}{z - x}\
&=2(z + x)-5
\end{align*}
$$
## Step3: Find the limit
Now, find $\lim_{z
ightarrow x}(2(z + x)-5)$. Substitute $z=x$ into $2(z + x)-5$.
$$
\begin{align*}
\lim_{z
ightarrow x}(2(z + x)-5)&=2(x + x)-5\
&=4x-5
\end{align*}
$$

# Answer:
$4x - 5$

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QUESTION IMAGE

Question

Explanation:

Step1: Substitute \(f(z)\) and \(f(x)\)

Step2: Factor and simplify

Step3: Find the limit

Answer: