Find and simplify the difference quotient (f(x + h) - f(x))/h, where h is not 0, for f(x) = x^2 - 5x + 2.

Step-by-step solution

1. 1

   Substitute $x+h$ into the whole rule, not just the first term. Every $x$ in $f$ must be replaced, including the one inside $-5x$:

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

   Expanding $(x+h)^{2}=x^{2}+2xh+h^{2}$ and distributing the $-5$:

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

2. 2

   Subtract $f(x)$ and keep the minus sign attached to every term. This is where most sign errors happen, so write the second bracket out before removing it:

   $f(x+h)-f(x)=\left(x^{2}+2xh+h^{2}-5x-5h+2\right)-\left(x^{2}-5x+2\right).$

   $=x^{2}+2xh+h^{2}-5x-5h+2-x^{2}+5x-2.$

3. 3

   Watch what cancels — and why. Every term of $f(x)$ that contains no $h$ disappears: $x^{2}-x^{2}=0$, $-5x+5x=0$, $2-2=0$. That always happens, and it is what guarantees the numerator is divisible by $h$:

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

4. 4

   Factor $h$ out of the numerator, then cancel it. Since $h\neq 0$ is given, the cancellation is legal:

   $\frac{2xh+h^{2}-5h}{h}=\frac{h(2x+h-5)}{h}=2x+h-5.$

5. 5

   Interpret and check the result. The answer $2x+h-5$ is the slope of the secant line through the points $(x,f(x))$ and $(x+h,f(x+h))$. Letting $h\to 0$ gives $2x-5$, which is exactly $f'(x)$ — a good confirmation. Numerical check at $x=1.7$, $h=0.03$: $\frac{f(1.73)-f(1.7)}{0.03}=-1.57$, and $2(1.7)+0.03-5=-1.57$.