Answer:
The values of a and b are [tex]a=\frac{1}{4}[/tex] and b = 1
Step-by-step explanation:
* Lets explain how to solve the equation
f(x) = {x² x ≤ 2
{ax³ + bx x > 2
* We need to find the values of a , b that make the function
differentiable at x = 2
- At first for f(x) to be continuous at x = 2, substitute x by two in the
the two expressions and equate them
∵ f(x) = x² at x ≤ 2 and f(x) = ax³ + bx at x > 2
∴ f(2) = (2)² = 4 ⇒ (1)
∴ f(2) = a(2)³ + b(2)
∴ f(2) = 8a + 2b ⇒ (2)
- Equate (1) and (2)
∴ 8a + 2b = 4 ⇒ (3)
* For f(x) to be differentiable when x = 2, the function must be
continuous when x = 2 and the one-sided derivatives must be
equal when x = 2
# Remember: If [tex]f(x)=ax^{b}[/tex] , then [tex]f'(x)=abx^{b-1}[/tex]
If [tex]f(x)=ax[/tex] , then [tex]f'(x)=a[/tex]
If [tex]f(x)=a[/tex] , then [tex]f'(x)=0[/tex]
∵ f(x) = x²
∴ f'(x) = 2x
- Substitute x by 2
∴ f'(2) = 2(2) = 4
∴ f'(2) = 4 ⇒ (4)
∵ f(x) = ax³ + bx
∴ f'(x) = 3ax² + b
- substitute x by 2
∴ f'(2) = 3a(2)² + b
∴ f'(2) = 12a + b ⇒ (5)
- Equate (4) and (5)
∴ 12a + b = 4 ⇒ (6)
* Now we have system of equations
8a + 2b = 4 ⇒ (3)
12a + b = 4 ⇒ (6)
- Multiply equation (6) by -2 to eliminate b
∴ -24 a - 2b = -8 ⇒ (7)
- Add equations (3) and (7)
∴ -16a = -4
- Divide both sides by -16
∴ a = [tex]\frac{1}{4}[/tex]
- substitute the value of a in equation (6)
∴ [tex]12(\frac{1}{4})+b=4[/tex]
∴ 3 + b = 4
- Subtract 3 from both sides
∴ b = 1
* The values of a and b are [tex]a=\frac{1}{4}[/tex] and b = 1
