Posted by Yoona on Monday, November 21, 2011 at 8:25am.
f(x) is both functions.. I dont know why the system didnt let me keep the spaces to show that
First off, stop misspelling "calculus".
Now, if a=2 and b=3, we have
f(x) = 2x^2 - 3x
f(1) = -1
But lim x->1- = |1-1| + 2 = 2
So, f is not continuous at x=1.
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If f is to be continuous, it needs to be continuous at x=1. It is continuous everywhere else already.
So, ax^2 - bx must = 2 at x=1
a - b = 2
So, there are any number of parabolas which will make f continuous at x=1.
6x^2 - 4x
-3x^2 + 5x
etc.
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Now, for f to be also differentiable, the slopes must match at x=1
The slope of |x-1| is -1 when approaching from the left.
ax^2 - (a-2)x must also have slope -1 at x=1
f'(x) = 2ax - (a-2)
f'(1) = 2a - a + 2 = -1
a = -3
So,
f(x) = |x-1| + 2 for x<1
f(x) = -3x^2 + 5x for x>=1
is both continuous and differentiable everywhere.
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