Where is f(x) = (x^2 – 4x – 12)/(x^2 – x – 6) not differentiable?
a) f(x) is not differentiable at x = 3
b) f(x) is not differentiable at x = 3 and x = -2
c) f(x) is not differentiable at x = -2
d) f(x) is differentiable for all x variables that are Real numbers (members of the Real number series)
2007-05-11 04:50:54 · 9 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
B. because this is where the function is not continuous. And if it's not continuous then it's not differentiable.
2007-05-11 04:55:11 · answer #1 · answered by Anonymous · 0⤊ 0⤋
I am not sure but I think that it is not differentiable where the denominator = 0, so the roots of x^2 - x - 6 are 3 and -2 so the answer would be b. Except for one thing, the numerator has a zero at -2 also so this equation simplifies to
(x + 2)(x - 6)/(x + 2)(x - 3) = (x - 6) / ( x - 3) so it is not differentiable at x = 3 only, therefore the answer is a.
2007-05-11 04:58:50 · answer #2 · answered by rscanner 6 · 0⤊ 0⤋
As a rational function, f(x) is differentiable for all x in its domain. So, wherever f(x) is defined, it will be differentiable. So, where is it defined?
It is defined for all x such that the denominator is not 0. So, to exclude those values, you set the numerator to 0, and solve the equation:
x^2-x-6=0
The quadratic formula gives us:
x1=[1+sqrt(1+4*6)]/2
=(1+5)/2
=3
x2=[1-sqrt(1+4*6)]/2
=(1-5)/2
=-2
So f(x) is defined, and hence differentiable, whenever x is not 3 or -2
This gives you B. as the correct answer.
2007-05-11 04:58:48 · answer #3 · answered by Yo 2 · 1⤊ 0⤋
B
(x^2-x-6) = (X-3)*(X+2)
2007-05-11 04:55:40 · answer #4 · answered by Grant d 4 · 0⤊ 0⤋
denominator is (x-3)(x+2), with roots x = 3 and -2.
f is not defined at x = 3 and -2 , so not continuous there, so not differentiable there.
So b is the best answer.
But a and c are good answers too.
2007-05-11 04:55:46 · answer #5 · answered by fcas80 7 · 0⤊ 0⤋
factor the top and bottom. the x+2 factor cancels out, leaving a hole in the graph at x = -2. not continuous there, so not differentiable. the x-3 factor doesn't cancel out, so you have a vertical asymptote at x = 3, also not differentiable.
2007-05-11 04:58:35 · answer #6 · answered by Philo 7 · 0⤊ 0⤋
b), this is where the denominator = 0, which as you should know, is a discontinuity:
Rule of Thumb;
Differentiability implies continuity, therefore a discontinuity implies it is not differentiable.
Anticipation:
Can't we factorise and simplify? No. This would make it a different function.
2007-05-11 04:59:01 · answer #7 · answered by qspeechc 4 · 0⤊ 0⤋
Simplified (x-6) / (x-3) x cannot = 3
2007-05-11 04:57:04 · answer #8 · answered by dwinbaycity 5 · 0⤊ 0⤋
Answer is (a)
f(x) = (x-6)(x+2)/((x-3)(x+2))
(x+2) in both numerator and denominator cancel out so..
f(x) = (x-6)/(x-3)
f'(x) = (x-3) - (x-6)/(x-3)^2 = 3/(x-3)^2
which is undefined at x=3.
2007-05-11 05:05:39 · answer #9 · answered by Anonymous · 0⤊ 0⤋