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
3 answers
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.
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
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
1 answer