let f(x)= bx^2+6x, if x< or equal to 2
ax^3, if x>2
find the values of a and b such that f(x)is differentiable at x=2
I need help with this question I know you have to set both functions equal to each other. but I don't know how to solve for a and b
f(x) =
bx^2+6x if x <= 2
ax^3 if x>2
So, we have
lim(x->2-) f(x) = 4b+12
lim(x->2+) f(x) = 8a
That means we need
4b+12 = 8a
Also, we need the derivatives to be equal:
2bx+6 -> 4b+6
3ax^2 -> 12a
And they need to be equal: 4b+6 = 12a
rearranging things a bit, we have
8a-4b = 12
12a-4b = 6
a = -3/2
b = -6
f(x) =
-6x^2+6x if x <= 2
-3/2 x^3 if x > 2
Check the graphs. There should be a smooth transition at x=2:
http://www.wolframalpha.com/input/?i=plot+y+%3D+-6x^2%2B6x%2C+y+%3D+-3%2F2+x^3+for+x+%3D+-1..3
To find the values of a and b such that the function f(x) is differentiable at x = 2, we need to ensure that both parts of the function are continuous and have the same derivative at x = 2.
First, let's set up the conditions for continuity and differentiability at x = 2.
Condition 1: f(x) is continuous at x = 2.
Since f(x) is defined differently for x ≤ 2 and x > 2, we need to calculate the left and right limits of f(x) as x approaches 2 from both sides.
For x ≤ 2:
lim (x→2-) f(x) = lim (x→2-) (bx^2 + 6x) = 4b + 12 ...(1)
For x > 2:
lim (x→2+) f(x) = lim (x→2+) (ax^3) = 8a ...(2)
To satisfy the condition of continuity at x = 2, equation (1) and (2) should be equal. Therefore, we have:
4b + 12 = 8a ...(3)
Condition 2: f(x) has the same derivative at x = 2.
For f(x) to be differentiable at x = 2, the derivatives of the two parts of the function should be equal at x = 2.
The derivative of the first part (x ≤ 2):
f'(x) = 2bx + 6
To find the derivative of the second part (x > 2), we take the derivative of ax^3:
f'(x) = 3ax^2
Now, let's evaluate the derivatives at x = 2:
For x ≤ 2:
f'(2) = 2b(2) + 6 = 4b + 6 ...(4)
For x > 2:
f'(2) = 3a(2)^2 = 12a ...(5)
To satisfy the condition of the same derivative at x = 2, equations (4) and (5) should be equal. Therefore, we have:
4b + 6 = 12a ...(6)
Now, we have two equations (3) and (6) with two unknowns (a and b). We can solve these equations to find the values of a and b.
From equation (3): 4b + 12 = 8a
From equation (6): 4b + 6 = 12a
Subtracting equation (6) from equation (3) gives:
4b + 12 - (4b + 6) = 8a - 12a
6 = -4a
Dividing both sides by -4 gives:
a = -6/4 = -3/2
Substituting the value of a into equation (6) gives:
4b + 6 = 12(-3/2)
4b + 6 = -18
4b = -24
b = -6
Therefore, the values of a and b that satisfy the conditions for f(x) to be differentiable at x = 2 are a = -3/2 and b = -6.