We want to get the difference quotient for the given function, we will get see that the difference quotient is equal to:
f'(x) = 2x + 5
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The difference quotient:</h3>
For a given function f(x), we define the difference quotient as:
![\lim_{h \to 0} \frac{f(x + h) - f(x)}{h}](https://tex.z-dn.net/?f=%5Clim_%7Bh%20%5Cto%200%7D%20%20%5Cfrac%7Bf%28x%20%2B%20h%29%20-%20f%28x%29%7D%7Bh%7D)
In this case, we have:
f(x) = x^2 + 5x + 6
Replacing that in the difference quotient we get:
![\lim_{h \to 0} \frac{(x + h)^2 + 5*(x + h) + 6 - x^2 - 5x - 6}{h}\\\\\lim_{h \to 0} \frac{x^2 + 2xh + h^2 + 5*x + 5*h + 6 - x^2 - 5x - 6}{h}\\\\\lim_{h \to 0} \frac{ 2xh + h^2 + 5*h }{h}\\](https://tex.z-dn.net/?f=%5Clim_%7Bh%20%5Cto%200%7D%20%20%5Cfrac%7B%28x%20%2B%20h%29%5E2%20%2B%205%2A%28x%20%2B%20h%29%20%2B%206%20-%20x%5E2%20-%205x%20-%206%7D%7Bh%7D%5C%5C%5C%5C%5Clim_%7Bh%20%5Cto%200%7D%20%20%5Cfrac%7Bx%5E2%20%2B%202xh%20%2B%20h%5E2%20%2B%205%2Ax%20%2B%205%2Ah%20%2B%206%20-%20x%5E2%20-%205x%20-%206%7D%7Bh%7D%5C%5C%5C%5C%5Clim_%7Bh%20%5Cto%200%7D%20%20%5Cfrac%7B%202xh%20%2B%20h%5E2%20%2B%205%2Ah%20%7D%7Bh%7D%5C%5C)
Now we can cancel the factor h to get:
![\lim_{h \to 0} 2x + h + 5 = 2x + 0 + 5 = 2x + 5](https://tex.z-dn.net/?f=%5Clim_%7Bh%20%5Cto%200%7D%20%202x%20%2B%20h%20%2B%205%20%3D%202x%20%2B%200%20%2B%205%20%3D%202x%20%2B%205)
So the difference quotient is equal to 2x + 5.
If you want to learn more about difference quotients, you can read:
brainly.com/question/4224465