<p>i know this is probably really easy to most of you but i need help.</p>
<li>let f be the function defined as follows.</li>
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<p>f(x)= abs value (x-1) + 2, for x < 1
ax^2 + bx, for x greater than or equal to 1, where a and b are constants</p>
<p>b. describe all values of a and b for which f is a continuous function.</p>
<p>c. for what values of a and b is f both continuous and differentiable?</p>
<p>Assuming f(x) = ABS(x-1) + 2 instead of ABS( (x-1) + 2) for x < 1 :</p>
<p>If x < 1, x-1 will always be < 0, so ABS(x-1) will be the same as (1-x). So, for x < 1 we can write f(x) in a simpler form
f(x) = 1 -x + 2 = 3 -x for x < 1</p>
<p>b. For f(x) to be continuous, it cannot have a 'break' in function value at x=1. We know that 3-x will equal 2 when x=1, so we need
ax^2 + bx = 2 when x=1,
or a + b = 2 for f(x) to be continuous.</p>
<p>c. For x < 1, f'(x) = -1
For x >= 1, f'(x) = 2ax + b and so f'(1) = 2a + b = a + (a+b) = a+2, if f(x) is continuous
For both derivatives to be equal, we need -1 = f(1) or -1 = a + 2
So a = -3 and b = 2-a = 5 .</p>