f (x)={cx+d for x≤2
{-x^2−cx for x>2
Let f be the function defined above, where c and d are constants. If f is differentiable at x=2, what is the value of c+d ?
-4, -2, 0, 2, 4
you need f and f' to be continuous
lim(x→2-) f'(x) = c
lim(x→2+) f'(x) = -4-c
So, you need c = -4-c => c = -2
lim(x→2-) f(x) = 2c+d = -4+d
lim(x→2+) f(x) = -4+4 = 0
so, -4+d = 0 => d=4
S, f(x) =
{-2x+4 for x≤2
{-x^2+2x for x>2
See the graphs at
https://www.wolframalpha.com/input/?i=plot+y%3D-2x%2B4%2C+2x-x%5E2
Oh, I see we're getting a little mathematical here. Well, let's crunch some numbers!
To determine if f is differentiable at x = 2, we need to check if the limit of the derivative exists as x approaches 2 from both sides.
For x ≤ 2, we have f(x) = cx + d. Taking the derivative with respect to x gives us f'(x) = c.
For x > 2, we have f(x) = -x^2 - cx. Taking the derivative with respect to x gives us f'(x) = -2x - c.
Now, let's check the limits:
Limit as x approaches 2 from the left (x ≤ 2):
lim(x→2-) f'(x) = lim(x→2-) c = c
Limit as x approaches 2 from the right (x > 2):
lim(x→2+) f'(x) = lim(x→2+) (-2x - c) = -4 - c
For f to be differentiable at x = 2, the limits from both sides must be equal. Therefore, we have:
c = -4 - c
Solving this equation, we find:
2c = -4
c = -2
Now, we need to find the value of d. We can use either equation for x ≤ 2 or x > 2 to solve for d. Let's use the equation for x ≤ 2:
f(2) = c * 2 + d
Since f is differentiable at x = 2, we also know that f(2-) = f(2+) = f(2). So we have:
c * 2 + d = f(2) = f(2-) = f(2+) = -2 * 2^2 - c * 2
-4 + d = -8 - 2c
Plugging in the value we found for c, we get:
-4 + d = -8 + 4
d = -8 + 4 + 4
d = 0
Finally, we can find the value of c + d:
c + d = -2 + 0
c + d = -2
So, the value of c + d is -2.
To determine the value of c+d, we need to ensure that the function f is differentiable at x=2.
For a function to be differentiable at a specific point, the function's left-hand and right-hand derivatives must be equal at that point.
First, let's find the left-hand derivative of f at x=2. We will differentiate the first part of the function f(x) = cx + d with respect to x:
f(x) = cx + d
Differentiating f(x) with respect to x gives us:
f'(x) = c
Now, let's find the right-hand derivative of f at x=2. We will differentiate the second part of the function f(x) = -x^2 - cx with respect to x:
f(x) = -x^2 - cx
Differentiating f(x) with respect to x gives us:
f'(x) = -2x - c
To find the value of c+d, we need to equate the left-hand derivative and the right-hand derivative at x=2 and solve for c+d:
c = -2(2) - c
3c = -4
c = -4/3
Now that we have the value of c, we can substitute it back into the first equation to solve for d:
-4/3 + d = -4/3 + d
d = 0
Therefore, the value of c+d is -4/3 + 0 = -4/3.
So, the answer is -4/3, which is not one of the given options.
To find the value of c+d, we need to determine if the function f(x) is differentiable at x=2.
To do this, we have to check if the left-hand limit and the right-hand limit of f(x) as x approaches 2 are equal, and if the derivative of the function at x=2 exists.
For x ≤ 2, f(x) = cx + d.
Taking the limit as x approaches 2 from the left-hand side:
lim(x→2-) (cx + d) = 2c + d
For x > 2, f(x) = -x^2 - cx.
Taking the limit as x approaches 2 from the right-hand side:
lim(x→2+) (-x^2 - cx) = -4 - 2c
For f(x) to be differentiable at x=2, the left-hand limit and the right-hand limit should be equal. Therefore:
2c + d = -4 - 2c
Now, let's find the value of c+d by solving the equation. We can rearrange the equation:
2c + 2c = -4 - d
4c = -4 - d
c = (-4 - d) / 4
Substituting c in the first equation:
2((-4 - d) / 4) + d = -4 - 2((-4 - d) / 4)
Simplifying this equation gives:
-4 - d + 2d = -4 + 2(4 + d) / 4
Simplifying further:
-4 + d = -4 + 8 + 2d
d = 8 - 4
d = 4
Substituting the value of d back into the equation for c:
c = (-4 - 4) / 4
c = -8 / 4
c = -2
Therefore, the value of c+d is -2 + 4 = 2.
Hence, the value of c+d is 2.