Let's assign the value of x as x = 2.
To create a function f that is both continuous and differentiable at x = 2, we can start by creating a basic function that satisfies these conditions. One such function is f(x) = x^2.
By calculating the derivative of f(x), we can check if it is also satisfied at x = 2.
f(x) = x^2
f'(x) = 2x
Now, let's evaluate the function and its derivative at x = 2.
f(2) = 2^2 = 4
f'(2) = 2(2) = 4
The function f(x) = x^2 satisfies the given conditions and is both continuous and differentiable at x = 2.
To prove the continuity and differentiability of f(x) = x^2 at x = 2 using limit definitions, we can apply the limit definitions of continuity and differentiability.
Continuity at x = 2:
lim(x→2) f(x) = f(2) = 4
Since the limit equals the value of the function at x = 2, f(x) = x^2 is continuous at x = 2.
Differentiability at x = 2:
lim(x→2) [f(x) - f(2)]/(x - 2) = lim(x→2) [(x^2 - 4)/(x - 2)]
By simplifying the equation,
lim(x→2) [(x + 2)] = 4
Since the limit exists and equals to 4, f(x) = x^2 is differentiable at x = 2.
Overall, f(x) = x^2 satisfies the given conditions and is both continuous and differentiable at x = 2.
As a challenge for classmates:
Given x = 3,
State the values of a new function and its derivative: f(3) = 5, and f'(3) = 2.
Challenge: Create a function g that is both continuous and differentiable at x = 3.
Without first creating a function, assign values to a function and its derivative for a particular value of x. For example, state that and .
Create a function f such that the function satisfies the given conditions and is both continuous and differentiable at that value of x. Write the function and describe how you found it.
Use the limit definitions of continuity at a point and differentiability at a point to prove that your function is both continuous and differentiable at that value of x.
As a challenge for your classmates, state values of a new function and its derivative at a particular value of x and ask them to create a function g such that the function is both continuous and differentiable at that value of x.
3 answers
Without first creating a function, assign values to a function and its derivative for a particular value of x. For example, state that and .
Create a function f such that the function satisfies the given conditions and is both continuous and differentiable at that value of x. Write the function and describe how you found it.
Use the limit definitions of continuity at a point and differentiability at a point to prove that your function is both continuous and differentiable at that value of x.
As a challenge for your classmates, state values of a new function and its derivative at a particular value of x and ask them to create a function g such that the function is both continuous and differentiable at that value of x.
Create a function f such that the function satisfies the given conditions and is both continuous and differentiable at that value of x. Write the function and describe how you found it.
Use the limit definitions of continuity at a point and differentiability at a point to prove that your function is both continuous and differentiable at that value of x.
As a challenge for your classmates, state values of a new function and its derivative at a particular value of x and ask them to create a function g such that the function is both continuous and differentiable at that value of x.
Let's assign the value of x as x = 2.
To create a function f that is both continuous and differentiable at x = 2, we can start by assigning arbitrary values to the function and its derivative at x = 2. Let's say f(2) = 5 and f'(2) = 3.
To find the function f, we can integrate f'(x) = 3 with respect to x to find the original function.
∫ 3 dx = 3x + C
Since we want f(2) to be equal to 5, we can substitute x = 2 into the equation:
3(2) + C = 5
6 + C = 5
C = -1
Therefore, the function f(x) that satisfies the conditions f(2) = 5 and f'(2) = 3 is:
f(x) = 3x - 1
Now, let's prove the continuity and differentiability of f(x) = 3x - 1 at x = 2 using limit definitions.
Continuity at x = 2:
lim(x→2) f(x) = f(2) = 5
Since the limit equals the value of the function at x = 2, f(x) = 3x - 1 is continuous at x = 2.
Differentiability at x = 2:
lim(x→2) [f(x) - f(2)]/(x - 2) = lim(x→2) [(3x - 1 - 5)/(x - 2)]
By simplifying the equation,
lim(x→2) [(3x - 6)/(x - 2)] = 0
Since the limit exists and equals to 0, f(x) = 3x - 1 is differentiable at x = 2.
Therefore, the function f(x) = 3x - 1 satisfies the given conditions and is both continuous and differentiable at x = 2.
As a challenge for classmates:
Given x = 4,
State the values of a new function and its derivative: f(4) = 2, and f'(4) = -1.
Challenge: Create a function g that is both continuous and differentiable at x = 4.
To create a function f that is both continuous and differentiable at x = 2, we can start by assigning arbitrary values to the function and its derivative at x = 2. Let's say f(2) = 5 and f'(2) = 3.
To find the function f, we can integrate f'(x) = 3 with respect to x to find the original function.
∫ 3 dx = 3x + C
Since we want f(2) to be equal to 5, we can substitute x = 2 into the equation:
3(2) + C = 5
6 + C = 5
C = -1
Therefore, the function f(x) that satisfies the conditions f(2) = 5 and f'(2) = 3 is:
f(x) = 3x - 1
Now, let's prove the continuity and differentiability of f(x) = 3x - 1 at x = 2 using limit definitions.
Continuity at x = 2:
lim(x→2) f(x) = f(2) = 5
Since the limit equals the value of the function at x = 2, f(x) = 3x - 1 is continuous at x = 2.
Differentiability at x = 2:
lim(x→2) [f(x) - f(2)]/(x - 2) = lim(x→2) [(3x - 1 - 5)/(x - 2)]
By simplifying the equation,
lim(x→2) [(3x - 6)/(x - 2)] = 0
Since the limit exists and equals to 0, f(x) = 3x - 1 is differentiable at x = 2.
Therefore, the function f(x) = 3x - 1 satisfies the given conditions and is both continuous and differentiable at x = 2.
As a challenge for classmates:
Given x = 4,
State the values of a new function and its derivative: f(4) = 2, and f'(4) = -1.
Challenge: Create a function g that is both continuous and differentiable at x = 4.
1 answer