f(x) = sqrt(x^2 + 0.0001)
At x = 0, which of the statements is true.
a)f is increasing
b)f is discontinuous
c)f has a horizontal tangent
d)f' is undefined
Answer is c but why?
f(x,y) = sqrt(x^2 + y)
g(x,y) = df/dx =
1/(2*sqrt[x^2 + y]) * 2x =
x/sqrt[x^2 + y])
For y>0 this is zero when x = 0.
But for y = 0 you find:
g(x,0) = 1
This means that Lim y --> 0 of g(0,y)=0
Because f(x) = sqrt(x^2 + 0.0001) is a continuous function, we know that it is not discontinuous at x = 0 (option b). Additionally, f'(x) is defined for all values of x, so it is not undefined (option d).
To determine if f has a horizontal tangent at x = 0, we need to find the derivative of f(x) and evaluate it at x = 0. Taking the derivative of f(x) = sqrt(x^2 + 0.0001) with respect to x, we use the chain rule:
f'(x) = (1/2) * (x^2 + 0.0001)^(-1/2) * (2x)
= x / sqrt(x^2 + 0.0001)
Evaluating this derivative at x = 0, we get:
f'(0) = 0 / sqrt(0^2 + 0.0001)
= 0 / sqrt(0.0001)
= 0 / 0.01
= 0
Since f'(0) = 0, this tells us that the slope of the tangent line to f(x) at x = 0 is 0. A horizontal tangent line has a slope of 0, so we can conclude that f(x) has a horizontal tangent at x = 0 (option c).
To determine if f(x) has a horizontal tangent at x = 0, we need to examine the derivative, f'(x), and evaluate it at x = 0.
Given that f(x) = sqrt(x^2 + 0.0001), we can find the derivative using the chain rule. Let's differentiate f(x) with respect to x:
f'(x) = (1/2)(x^2 + 0.0001)^(-1/2) * 2x = x/(sqrt(x^2 + 0.0001))
Now, we need to evaluate the derivative at x = 0:
f'(0) = 0/(sqrt(0^2 + 0.0001)) = 0/0.01 = 0
As we can see, the derivative f'(0) exists and equals 0. This means that the slope of the tangent line to the graph of f(x) at x = 0 is 0.
A horizontal tangent occurs when the slope of the tangent line is 0, which is the case here. Therefore, the correct statement is c) f has a horizontal tangent at x = 0.
To determine which statement is true at x = 0 for the function f(x) = sqrt(x^2 + 0.0001), we can find the derivative of the function and examine its behavior.
The derivative of f(x) with respect to x, denoted as f'(x), can be found using the chain rule. Let's denote g(x,y) = sqrt(x^2 + y), where y = 0.0001. We can differentiate g(x,y) with respect to x.
g(x,y) = sqrt(x^2 + y)
g'(x,y) = (1/2) * (x^2 + y)^(-1/2) * (2x)
g'(x,y) = x / sqrt(x^2 + y)
Now, let's evaluate the derivative at x = 0, considering y > 0 and y = 0.
For y > 0, when x = 0, we have:
g'(0,y) = 0 / sqrt(0^2 + y)
g'(0,y) = 0
This tells us that the derivative is zero at x = 0 when y > 0, indicating a horizontal tangent.
However, for y = 0, we have:
g'(0,0) = 0 / sqrt(0^2 + 0)
g'(0,0) = 0/0
When evaluating the derivative at y = 0, we find an indeterminate form (0/0).
To investigate further, we can take the limit as y approaches 0:
lim (y->0) g'(0,y)
Taking the limit helps determine the behavior of the derivative at x = 0 when y = 0.
lim (y->0) g'(0,y) = lim (y->0) 0
lim (y->0) g'(0,y) = 0
So, the limit of the derivative as y approaches 0 is 0.
From this, we can conclude that at x = 0:
- For y > 0, the derivative is 0, indicating a horizontal tangent.
- For y = 0, the limit of the derivative is 0.
Therefore, the statement that is true at x = 0 is that f has a horizontal tangent (option c).