Given the function f(x) = 1/(sqrt x)
a. Does f(x) exist at x=0?
b. Does f'(x) exist at x=0?
c. Explain why integral (from 0->1) f(x)dx does exist.
THANK YOU.
a) no, one is not allowed to have zero in the denominator.
b. f(x)=1/sqrt(x)
f'= -1/(2 x-3/2) yes it exists
c. if an endpoint does not exist, one cannot integrate it.
Thank you so much! but are you sure for part c.?
so idk the answer
a. To determine if f(x) exists at x=0, we need to evaluate the function at that point. In this case, substitute x=0 into f(x) to find f(0). We have:
f(0) = 1/(√0)
However, the square root of 0 (√0) is not defined, as any number multiplied by itself equals 0. Therefore, f(x) does not exist at x=0.
b. To determine if f'(x) exists at x=0, we need to find the derivative of f(x) and evaluate it at x=0. Let's differentiate f(x):
f(x) = 1/(√x)
Using the power rule for differentiation, we have:
f'(x) = -(1/2) * (x)^(-3/2)
Now, let's evaluate f'(x) at x=0:
f'(0) = -(1/2) * (0)^(-3/2)
Here, we also encounter a problem. The expression (0)^(-3/2) is undefined, as any number raised to the power of -3/2 would involve dividing by zero. Therefore, f'(x) does not exist at x=0.
c. To determine if the integral of f(x) from 0 to 1 exists, we need to evaluate the definite integral:
∫[0,1] f(x) dx
The function f(x) = 1/(√x) is defined and continuous for x > 0. Therefore, we can integrate f(x) over the interval [0,1].
To compute the integral, we need to find the antiderivative of f(x). By applying the power rule for integration, we get:
∫ f(x) dx = 2√x + C
Now, let's evaluate the integral from 0 to 1:
∫[0,1] f(x) dx = [2√x] from 0 to 1
To find the integral at x=1, substitute 1 in for x:
[2√1] = 2
To find the integral at x=0, substitute 0 in for x:
[2√0] = 0
Therefore, the integral of f(x) from 0 to 1 exists and evaluates to 2.