Asked by Anonymous
Piecewise function problem.
Let f(x)={ax^2+1/3, x is greater than or equal to 1; bx-10/3, x<1. If the function is differentiable, find the sum of a+b.
Let f(x)={ax^2+1/3, x is greater than or equal to 1; bx-10/3, x<1. If the function is differentiable, find the sum of a+b.
Answers
Answered by
Steve
f(x) =
ax^2 + 1/3 for x >= 1
bx - 10/3 for x < 1
so, we need f(1) to be consistent. That is,
a + 1/3 = b - 10/3
we also need f'(1) to be consistent:
2a = b
so, a + 1/3 = 2a - 10/3
a = 11/3
b = 22/3
a+b = 11
ax^2 + 1/3 for x >= 1
bx - 10/3 for x < 1
so, we need f(1) to be consistent. That is,
a + 1/3 = b - 10/3
we also need f'(1) to be consistent:
2a = b
so, a + 1/3 = 2a - 10/3
a = 11/3
b = 22/3
a+b = 11
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