Why does $[f'(a)(a+\Delta x-a)+b]-[f'(a)(a-a)+b]=f'(a)\Delta x$? If I try to calculate this I get $f'(a)(\Delta x-1) \neq f'(a)\Delta x$.
Why does $[f'(a)(a+\Delta x-a)+b]-[f'(a)(a-a)+b]=f'(a)\Delta x$
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algebra-precalculus
2 Answers
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$$\begin{align}[f'(a)(a+\Delta x-a)+b]-[f'(a)(a-a)+b]&= [f'(a)(\Delta x)+b]-[f'(a)\times 0+b] \\ &= f'(a)(\Delta x)+b-b \\ &= f'(a)(\Delta x)\end{align} $$
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Some of the terms cancel. Look:
$$\begin{align} [f'(a)(\color{red}{a}+\Delta x\color{red}{-a})+b]-[f'(a)\color{blue}{(a-a)}+b]&= [f'(a)(\Delta x)+b]-[f'(a)\cdot \color{blue}{0}+b] \\ &= f'(a)(\Delta x) \color{green}{+b-b} \\ &= f'(a)(\Delta x).\end{align} $$