Let f(x)=
sin(pi(x)) , 0 is less than or equal to x is less than or equal to 1
a(x)+b , 1 is less than x is less than or equal to 2
Find the values of a and b such that f(x) is differentiable at x=1
a=-pi b=pi?
rephrasing your function:
Let f(x)=
sin(pi(x)) , 0 ≤ x ≤ 1
a(x)+b , 1 ≤ x ≤ 2
f ' (x) = πcos(πx)
f ' (1) = -1
since x = 1 lies in the domain of the first part of the f(x), and πcos(π) = -1, the values of a and b don't enter the picture.
you need f(x) to be continuous and the limits of f'(x) to be the same on both sides at x=1
f(x) = sin(πx) for x <= 1 and so f(1) = sinπ = 0
so, f'(x) = πcos(πx) and f'(1) = πcosπ = -π
taking the limit from the right as x->1+,
f(x) = ax+b, so a+b=0
f'(x) = a so a=-π
f(x) = -πx+π for x > 1
You can see from the graphs that they line up for a continuous derivative at x=1
https://www.wolframalpha.com/input/?i=plot+y%3Dsin(%CF%80x),+y%3D-%CF%80x%2B%CF%80
To determine the values of a and b such that the function f(x) is differentiable at x=1, we need to ensure that the left and right limits of f(x) are equal at x=1.
First, let's consider the left limit of f(x) as x approaches 1 from the left side:
lim(x→1-) f(x) = lim(x→1-) sin(πx).
Since π is a constant and x is approaching 1 from the left side, πx will approach π. Therefore, we can rewrite the left limit as:
lim(x→1-) f(x) = sin(π).
Now, let's consider the right limit of f(x) as x approaches 1 from the right side:
lim(x→1+) f(x) = lim(x→1+) (a(x) + b).
To ensure that the function is differentiable at x=1, the left and right limits must be equal:
lim(x→1-) f(x) = lim(x→1+) f(x).
Substituting the values of the left and right limits mentioned above:
sin(π) = lim(x→1+) (a(x) + b).
Since the value of sin(π) is equal to 0, we have:
0 = lim(x→1+) (a(x) + b).
Now, let's evaluate the right limit:
lim(x→1+) (a(x) + b) = a + b.
Since the right limit equals 0, we can conclude that:
a + b = 0.
To find the values of a and b, we need to find a relationship between them that satisfies the equation a + b = 0. There are various possible solutions, but one simple solution is to let a = 1 and b = -1. This way, a + b equals zero, satisfying the equation.
Therefore, the values of a and b that make f(x) differentiable at x=1 are a = 1 and b = -1.