Decide if the following function f(x) is differentiable at x=0. Try zooming in on a graphing calculator, or calculating the derivative f'(0) from the definition.
f(x) = x^4sin(2/x),
if x is not equal to 0,
and
f(x) = 0
if x = 0.
If it is differentiable, what is the derivative? (If it isn't, enter dne.)
f'(0) = ________
I graphed it, it did what I expected. You graph it near zero.
f'=4x^3 sin(2/x)-x^4 cos(2/x)*2/x^2
f'(0)=0 YOu need to know cos(2/x) is a max of 1, a min of -1, either way, that times 0 is zero.
Thank you very much!
Well, I'm not a graphing calculator, but I can still try to answer your question with a little bit of humor! Let's take a look at this function.
The function f(x) = x^4sin(2/x) is quite the mouthful, isn't it? It's like trying to pronounce a tongue twister while doing calculus. But fear not, because I'm here to make it a little less daunting!
Now, to determine if this function is differentiable at x = 0, we need to check whether the derivative exists. In other words, we need to see if the slope of the function is well-behaved at that point.
When we look at the definition of the function, we see that f(x) = 0 if x = 0. That suggests that the function is pretty well-behaved at that point. But let's not jump to conclusions just yet!
To find the derivative f'(0), we need to take its definition. Now, brace yourself because we're diving into some limit math here. Are you ready?
f'(0) = lim (h -> 0) [f(h) - f(0)] / h
So if you calculate this limit and find that it exists, then congratulations! The function is differentiable at x = 0. But of course, if it doesn't exist, then the answer would be "dne" (does not exist).
Remember, math can be funny sometimes, especially when you throw in some clowns like me! I hope I was able to give you a little smile while explaining all this. Good luck with your calculations!
To determine if the function f(x) = x^4sin(2/x) is differentiable at x = 0, we need to check if the limit of the derivative f'(x) exists as x approaches 0.
First, let's calculate the derivative of f(x), by applying the product rule and chain rule:
f'(x) = d/dx(x^4sin(2/x))
= 4x^3sin(2/x) + x^2cos(2/x) * d/dx(2/x)
= 4x^3sin(2/x) - 2x^2cos(2/x) / x^2
Now, we need to find the limit of f'(x) as x approaches 0. Let's simplify the expression:
lim(x->0) (4x^3sin(2/x) - 2x^2cos(2/x) / x^2)
Since x*sin(2/x) is bounded by [-1,1] and x*cos(2/x) approaches 0 as x approaches 0, we can use the Squeeze Theorem to find the limit:
lim(x->0) (4x^3sin(2/x) - 2x^2cos(2/x) / x^2)
= lim(x->0) (4x * sin(2/x) - 2cos(2/x)) / x
Breaking it down further, we have two separate limits to evaluate:
lim(x->0) (4x * sin(2/x)) / x and lim(x->0) (-2cos(2/x)) / x
For the first limit, we can divide both the numerator and denominator by x:
lim(x->0) (4 * sin(2/x) / 1) = 4 * sin(2/x)
For the second limit, we can use the fact that cos(2/x) is bounded by [-1,1]:
lim(x->0) (-2cos(2/x)) / x
= lim(x->0) -2 / x (cos(2/x) / 1)
= lim(x->0) -2 / x
As x approaches 0, the limit of both limits is not finite, which means the limit of f'(x) as x approaches 0 does not exist. Therefore, the function f(x) is not differentiable at x = 0.
Hence, the answer to the question "If it is differentiable, what is the derivative?" is "dne."