Let f(x) = 7sin(x)
Part A) |f'(x)|

Question

Let f(x) = 7sin(x)
Part A) |f'(x)| <= __?__
Part B) By the Mean Value Theorem, |f(a) - f(b)| <= __?__ |a - b|

anonymous · Answer

$f'(x)=7\cos x \implies |f'(x)|\le7$. By the mean value theorem, we have $f'(c)=\frac{f(a)-f(b)}{b-a}$. We can generalize this to any interval in the domain of f. We can then find that $|f(a)-f(b)|\le 7|a-b|$, since $|f'(c)|\le7$ for all c in the domain of f.

anonymous · Answer

A = 7 cos x

lgg23 · Answer

I have a question. I know how to get $$f \prime (x) = 7\cos (x)$$, but how does $$f \prime (x) = 7\cos x$$ result in $$\left| f \prime (x)  \right| \le 7$$?

anonymous · Answer

I'm sure you know that  $-1\le \cos x \le 1$. This is the same as writing $|\cos x|\le1$, then you just multiply both sides by $7$.

lgg23 · Answer

Oh yeah, that's right. I forgot. Thanks.

anonymous · Answer

You're welcome!