If f(x)={ax3−6x, x≤1bx2+2, x>1 , find the values of a and b that will make f(x) differentiable at x = 1.
To make f(x) differentiable at x = 1, we need the left-hand derivative to equal the right-hand derivative at that point.
Let's find the left-hand derivative, denoted by f'(1-) and the right-hand derivative, denoted by f'(1+), and equate them.
To find f'(1-), we consider the function f(x) for x ≤ 1:
f(x) = ax^3 - 6x
To find f'(x) for x ≤ 1, we differentiate f(x) with respect to x:
f'(x) = 3ax^2 - 6
Evaluating f'(1-) by substituting x = 1:
f'(1-) = 3a(1)^2 - 6
= 3a - 6
Now let's find f'(1+), considering the function f(x) for x > 1:
f(x) = bx^2 + 2
To find f'(x) for x > 1, we differentiate f(x) with respect to x:
f'(x) = 2bx
Evaluating f'(1+) by substituting x = 1:
f'(1+) = 2b(1)
= 2b
Since f(x) is continuous at x = 1, we have:
f(1-) = f(1+)
Substituting the expressions for f(1-) and f(1+), we have:
a(1)^3 - 6(1) = b(1)^2 + 2
This simplifies to:
a - 6 = b + 2
a - b = 8
Now, equating the left-hand derivative and the right-hand derivative, we have:
f'(1-) = f'(1+)
3a - 6 = 2b
Using the equation a - b = 8, we can solve for a and b simultaneously:
3a - 6 = 2(a - 8)
Expanding and simplifying:
3a - 6 = 2a - 16
a = -10
Substituting the value of a into a - b = 8:
-10 - b = 8
b = -18
Therefore, the values of a and b that will make f(x) differentiable at x = 1 are a = -10 and b = -18.
To determine the values of a and b that will make f(x) differentiable at x = 1, we need to ensure that the function is both continuous and has the same rate of change from the left and right of x = 1.
First, let's check for continuity at x = 1. The function must have the same value from both the left and right of x = 1.
For x ≤ 1:
f(x) = ax^3 - 6x
For x > 1:
f(x) = bx^2 + 2
To guarantee continuity at x = 1, we need the left-hand limit to be equal to the right-hand limit.
Left-hand limit (LHL):
LHL = lim(x -> 1-) f(x) = lim(x -> 1-) (ax^3 - 6x) = a(1)^3 - 6(1) = a - 6
Right-hand limit (RHL):
RHL = lim(x -> 1+) f(x) = lim(x -> 1+) (bx^2 + 2) = b(1)^2 + 2 = b + 2
For continuity at x = 1, we require LHL = RHL:
a - 6 = b + 2 (equation 1)
Now, let's check for the same rate of change from the left and right of x = 1. The derivative of f(x) must be defined and equal from both sides.
Derivative of f(x) for x ≤ 1:
f'(x) = d/dx (ax^3 - 6x) = 3ax^2 - 6
Derivative of f(x) for x > 1:
f'(x) = d/dx (bx^2 + 2) = 2bx
To ensure the same rate of change at x = 1, we need the derivatives to be equal:
Derivative equality:
3ax^2 - 6 = 2bx (equation 2)
Now, we have two equations (equation 1 and equation 2) with two unknowns (a and b). We can solve these equations simultaneously to find the values for a and b.
From equation 1:
a - 6 = b + 2 => a = b + 8 (equation 3)
Substitute equation 3 into equation 2:
3(b + 8)(1)^2 - 6 = 2b
3(b + 8) - 6 = 2b
3b + 24 - 6 = 2b
b + 18 = 2b
18 = b (substituting b = 18 into equation 3)
a = 18 + 8 = 26
Therefore, the values of a = 26 and b = 18 will make f(x) differentiable at x = 1.
In order for f(x) to be differentiable at x = 1, the following conditions must be met:
1. f(x) must be continuous at x = 1.
2. The left-hand derivative (LHD) of f(x) at x = 1 must equal the right-hand derivative (RHD) of f(x) at x = 1.
First, let's check the continuity of f(x) at x = 1.
Consider the limit as x approaches 1 from the left side:
lim (x→1-) f(x) = lim (x→1-) (ax^3 - 6x)
Substituting x = 1 into the expression, we get:
lim (x→1-) f(x) = a(1)^3 - 6(1) = a - 6
Consider the limit as x approaches 1 from the right side:
lim (x→1+) f(x) = lim (x→1+) (bx^2 + 2)
Substituting x = 1 into the expression, we get:
lim (x→1+) f(x) = b(1)^2 + 2 = b + 2
For f(x) to be continuous at x = 1, the following condition must hold:
a - 6 = b + 2
Next, let's check the equality of the LHD and RHD of f(x) at x = 1.
LHD of f(x) at x = 1:
lim (x→1-) [f(x) - f(1)] / (x - 1) = lim (x→1-) [(ax^3 - 6x) - (a(1)^3 - 6(1))] / (x - 1)
= lim (x→1-) [(ax^3 - 6x) - (a - 6)] / (x - 1)
= lim (x→1-) (ax^3 - 6x - a + 6) / (x - 1)
Using L'Hôpital's rule to evaluate the limit:
= lim (x→1-) [3ax^2 - 6] / 1
= 3a - 6
RHD of f(x) at x = 1:
lim (x→1+) [f(x) - f(1)] / (x - 1) = lim (x→1+) [(bx^2 + 2) - (b(1)^2 + 2)] / (x - 1)
= lim (x→1+) [(bx^2 + 2) - (b + 2)] / (x - 1)
= lim (x→1+) (bx^2 - b) / (x - 1)
Using L'Hôpital's rule to evaluate the limit:
= lim (x→1+) [2bx] / 1
= 2b
For f(x) to be differentiable at x = 1, the following condition must hold:
3a - 6 = 2b
We now have two equations:
a - 6 = b + 2 (from the continuity condition)
3a - 6 = 2b (from the differentiability condition)
Simplifying the first equation, we get:
a - b = 8 (Equation 1)
Simplifying the second equation, we get:
3a - 2b = 6 (Equation 2)
We can now solve these two equations simultaneously.
Multiplying Equation 1 by 3, we get:
3a - 3b = 24 (Equation 3)
Subtracting Equation 2 from Equation 3, we eliminate the variable "a":
(3a - 3b) - (3a - 2b) = 24 - 6
-3b + 2b = 18
-b = 18
b = -18
Substituting the value of b into Equation 1, we can solve for a:
a - (-18) = 8
a + 18 = 8
a = -10
Therefore, the values of a and b that make f(x) differentiable at x = 1 are:
a = -10 and b = -18.