1. ## derivatives

f(x)=ax^5 when x<f
f(x)=bx-4 when x>=f

for f(x) to be differentiable when x=f what should a and b be?

tysm

2. ## Re: derivatives

f(x)=ax^5 when x<f
f(x)=bx-4 when x>=f
let's call the x-value of interest $c$ instead of $f$, so as not to confuse $f$ and $y = f(x)$ ...

first, the function, $f(x)$, has to be continuous at $\displaystyle x=c \implies \lim_{x \to c^-} f(x) = \lim_{x \to c^+} f(x) \implies ac^5 = bc-4$

second, the function, $f'(x)$, must also be continuous at $\displaystyle x=c \implies \lim_{x \to c^-} f'(x) = \lim_{x \to c^+} f'(x) \implies 5ac^4 = b$

can you finish?

3. ## Re: derivatives

Originally Posted by skeeter
let's call the x-value of interest $c$ instead of $f$, so as not to confuse $f$ and $y = f(x)$ ...

first, the function, $f(x)$, has to be continuous at $\displaystyle x=c \implies \lim_{x \to c^-} f(x) = \lim_{x \to c^+} f(x) \implies ac^5 = bc-4$

second, the function, $f'(x)$, must also be continuous at $\displaystyle x=c \implies \lim_{x \to c^-} f'(x) = \lim_{x \to c^+} f'(x) \implies 5ac^4 = b$

can you finish?
should the answer be in terms of f ?

4. ## Re: derivatives

Originally Posted by DiscreteMathHelp
should the answer be in terms of f ?
yes ... change $c$ to $f$ when you are done.