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