# How do you find the difference quotient of f, that is, find ((f(x+h)-f(x))/h), h != 0 for f(x)=x^2-5x+7?

Dec 14, 2017

see explanation

#### Explanation:

If I understand the question correctly, you have to start by substituting $\left(x + h\right)$ wherever you see $x$ in your original function definition:

${\left(x + h\right)}^{2} - 5 \left(x + h\right) + 7$

multiplying out, you have: ${x}^{2} + 2 x h + {h}^{2} - 5 \left(x + h\right) + 7$

$= {x}^{2} + 2 x h + {h}^{2} - 5 x - 5 h + 7$

So, substitute this as the value of $f \left(x + h\right)$ in the definition of the difference quotient.

$\frac{f \left(x + h\right) - f \left(x\right)}{h}$

$= \frac{\left({x}^{2} + 2 x h + {h}^{2} - 5 x - 5 h + 7\right) - \left({x}^{2} - 5 x + 7\right)}{h}$

...this simplifies to:

$\frac{2 x h + {h}^{2} - 5 h}{h}$

...now, since this is calculus, the inevitable next step is to find the limit of this function as h -> 0. For this, you can't have h in the denominator. But we can do a little factoring kung-fu. The equation above can be re-written:

$\frac{h \left(2 x + h - 5\right)}{h} = 2 x - 5 + h$

...and the limit of this function, as h approaches 0, is :

$2 x - 5$

...which is the derivative of the original function ${x}^{2} - 5 x + 7$

GOOD LUCK