Question

let f be the function defined by
|x-1|+2 for X<1
f(x)=
ax^2-Bx, for X>or equal to 1. where a and b are constants

a)if a=2 and b=3 is f continious for all x? justify your answer

b)describe all the values of a and b for which f is a continious function

c) For what values of a and b is f both continious and differentiable?

Answer 1

f(x) is both functions.. I don't know why the system didn't let me keep the spaces to show that

Answer 2

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.
__________________________

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.