Question

f(x) = -2x^2+4x for x <0 and 8x^2-3 for x greater or equal to zero
what would the difference quotient's be for lim x->0- and lim x-> 0+?

Answer 1

Function is defined as:
\[f(x)=-2x^2+4x, for\ x<0\]
\[f(x)=8x^2-3. for\ x\ge 0\]
for limit 0-, f(x) defined for x<0
\[\large \lim_{x \to 0^{-}} (-2x^2+4x)\]
for limit 0+, f(x) defined for \(x\ge 0\)
\[\large \lim_{x \to 0^{+}} (8x^2-3)\]

Answer 2

@darkmare do you get this?

Answer 3

yes I understand that for the limit of each one that those are used but the questions asks the lim of those values for f prime (0) so those are not the answers they are looking for apparently

Answer 4

f(x)=−2x2+4x,for x<0
f(x)=8x2−3.for x≥0
for limit 0-, f(x) defined for x<0
lim x→0−(−2x2+4x)= 0

for limit 0+, f(x) defined for x≥0
lim x→0+(8x2−3)=-3

Answer 5

\[\large f'(x)=\lim_{h \to 0^-}\frac{f(x-h)-f(0)}{x-h-x}\]
to evaluate f'(0), from left hand side ( or to check differentiability)
put x=0 in this
\[\large f'(x)=\lim_{h \to 0^-}\frac{-2(x-h)^2+4(x-h)-(-2x^2+4x)}{x-h-x}\]

now put x=0 in this, for 0+, use f(x)=8x^2-3

Answer 6

do you get it ? @darkmare

Answer 7

yes I do understand that, it is still not accepting it as an answer so I will have to look into it further, perhaps simplify further

Answer 8

yeah, you have to simplify it further.