At x = 3, the function given by f(x) = { x² , x<3} ; {6x-9 , x ≥ 3} is
a. continuous but not differentiable.
b. differentiable but not continuous.
c. neither continuous nor differentiable.
d. both continuous and differentiable.
Well, isn't this function coming at us with a twist! At x = 3, it decides to change its outfit. Before x = 3, it's all about the x² fashion statement, but after x = 3, it opts for the 6x - 9 ensemble.
Now, let's tackle this question like we tackle fashion emergencies. At x = 3, both parts of the function exist and are well-defined. This means that the function is continuous at x = 3. So, we can confidently eliminate options (c) and (d).
But wait, there's more! When it comes to differentiability, this function has a smooth transition at x = 3. Both parts have well-defined derivatives, meaning option (b) can leave the runway.
Now, we're left with option (a). This seems to be the right fit for our function. It's continuous at x = 3, but it changes its style, so it's not differentiable.
So, in the world of functions, our little fashionista is "continuous but not differentiable." Option (a) is our winner!
To determine if the function is continuous at x = 3, we need to check if the left-hand limit and the right-hand limit of the function exist and are equal at x = 3.
For x < 3, the function is given by f(x) = x². Therefore, the left-hand limit as x approaches 3 can be found by substituting 3 into the function:
lim(x→3-) f(x) = lim(x→3-) x² = 3² = 9.
For x ≥ 3, the function is given by f(x) = 6x - 9. Therefore, the right-hand limit as x approaches 3 can be found by substituting 3 into the function:
lim(x→3+) f(x) = lim(x→3+) (6x - 9) = (6 * 3) - 9 = 9.
Since the left-hand limit (9) is equal to the right-hand limit (9), the function is continuous at x = 3.
To determine if the function is differentiable at x = 3, we need to check if the derivative exists at x = 3.
For x < 3, the derivative of f(x) = x² can be found by differentiating using the power rule:
f'(x) = d/dx (x²) = 2x.
For x ≥ 3, the derivative of f(x) = 6x - 9 can be found by differentiating using the power rule:
f'(x) = d/dx (6x - 9) = 6.
Since both derivatives exist and are equal at x = 3 (2x = 6), the function is differentiable at x = 3.
Therefore, the correct answer is d. both continuous and differentiable.
To determine whether the given function is continuous and differentiable at x = 3, we need to check the conditions for continuity and differentiability separately in each of the two defined parts of the function at x = 3.
First, let's consider the function f(x) = x² for x < 3. In this region, f(x) is a simple quadratic function, which is continuous and differentiable for all values of x. Therefore, f(x) is continuous and differentiable at x = 3 from the left side.
Next, let's consider the function f(x) = 6x - 9 for x ≥ 3. In this region, f(x) is a linear function, which is also continuous and differentiable for all values of x. Hence, f(x) is continuous and differentiable at x = 3 from the right side.
Since the function is continuous from both the left and right sides at x = 3, and both parts of the function have derivatives at x = 3, we can conclude that the function as a whole is continuous and differentiable at x = 3.
Therefore, the correct answer is (d) both continuous and differentiable.
lim(x→3-) = 3^2 = 9
lim(x→3+) = 6*3-9 = 9
so f(x) is continuous
now check f' and see whether it is also the same from both sides.