Find the values of the constants 𝑎, 𝑏 and 𝑐 for which the following function is differentiable. (Give 𝑎 and 𝑏 in terms of 𝑐.)
𝑓(𝑥)={𝑐𝑥^2+4𝑥+1, 𝑥≥1 & 𝑎𝑥+𝑏, 𝑥<1
I found a = 2c+4
b = ??? Seems like this is pretty easy, however I tend to overshoot problems so I can't figure this out, need some help, thank you!
for x < 1,
f(x) = ax+b
f'(x) = a
For x > 1,
f(x) = cx^2+4x+1
f'(x) = 2cx
So, as x -> 1 from the left
f(1) = a+b
f'(1) = a
As x -> 1 from the right
f(1) = c+5
f'(1) = 2c+4
So, we need
a+b = c+5
a = 2c+4
2c+4 + b = c+5
b = 1-c
So, for example, if c=3,
a = 10, b = -2
check:
f(x) = 3x^2 + 4x+1 for x > 1
f(x) = 10x-2 for x < 1
At x=1,
f(x) = 8, f'(x) = 10
for both sides.
To find the value of constant b in terms of c, we need to consider the differentiability condition at x = 1.
For a function to be differentiable at a particular point, the two sides of the function should have the same value and the same derivative at that point.
Let's calculate the value of f(x) at x = 1 for both cases:
- For x ≥ 1: f(x) = cx^2 + 4x + 1
Substituting x = 1, we have f(1) = c(1)^2 + 4(1) + 1 = c + 4 + 1 = c + 5.
- For x < 1: f(x) = ax + b
Substituting x = 1, we have f(1) = a(1) + b = a + b.
Since the function should be continuous at x = 1, the values on both sides should be equal. Therefore, we have:
c + 5 = a + b
We already know that a = 2c + 4. Substituting this value in the equation above, we get:
c + 5 = (2c + 4) + b
Rearranging the equation, we find:
b = c + 5 - 2c - 4
b = -c + 1
Therefore, the value of the constant b in terms of c is b = -c + 1.