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Help me to find these unknown values?
1. Let p and q be real numbers and let f be the function defined by the piece function:

f(x) = 1+2p(x-1)+(x-1)^2 [x less or equal to 1]
qx+p [x greater than 1]

a. Find the values of p and q for which f is differentiable at x=1.
b. Using your values of p and q, is f" continuous?

2007-12-09 04:29:21 · 2 answers · asked by Anonymous in Science & Mathematics Mathematics

2 answers

f(x) = 1 + 2p(x - 1) + (x - 1)^2
f(1) = 1
f '(x) = 2p + 2(x - 1)

f(x) = qx + p
f(1) = q + p = 1
q = 1 - p
f '(x) = q

Set the two expressions for f '(1) equal to eachother:
2p = q
But q = 1 - p:
2p = 1 - p
3p = 1
p = 1/3
q = 2/3
The definition for f becomes:
f(x) = 1 + (2/3)(x - 1) + (x - 1)^2 [x f(x) = (2/3)x + 1/3 [x > 1]
Both parts of this definition are continuous and have the same slope at x = 1, hence f ' is continuous.

2007-12-09 12:12:12 · answer #1 · answered by jsardi56 7 · 0 0

a(x-a million)(x+2) + b(x-3)(x+2) + cx = x²-5x+3 a(x^2 + x - 2) + b(x^2 - x - 6) + cx = x^2 - 5x + 3 ax^2 + ax - 2a + bx^2 - bx - 6b + cx = x^2 - 5x + 3 (a + b)x^2 + (a - b + c)x -2(a + 3b) = x^2 - 5x + 3 a + b = a million a - b + c = -5 -2a - 6b = 3 2a + 2b = 2 -4b = 5 b = -5/4 a = 5/4 + a million a = 9/4 c = -5 - 9/4 - 5/4 c = -17/2

2016-12-17 12:22:01 · answer #2 · answered by wingert 4 · 0 0

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