Question

Find all values of c such that f is continuous on (-[infinity], [infinity]). f(x) = { 3 - x^2 x less than or equal to c x, x > ca. c = -1 + squareroot 13/2, -1 - squareroot 13/2.b. c = 0. c. c = -1 = squareroot 13/2.d. c = 2.e. c = -1 + squareroot 13/2, 1 - squareroot 13/2.

Answer 1

Answer:

a) $c=\frac{-1+\sqrt{13}}{2}$ and $c=\frac{-1-\sqrt{13}}{2}$

Step-by-step explanation:

The idea for the solution of this equation is to find the value of c where both parts of the piecewise-defined function are the same. So we need to take the parts of the function and set them equal to each other, so we get:

$3-x^{2}=x$

and then solve for x. We move everything to one side of the equation so we get:

$x^{2}+x-3=0$

and we use the quadratic formula:

$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

and we substitute:

$x=\frac{-1\pm \sqrt{(1)^2-4(1)(-3)}}{2(1)}$

and solve

$x=\frac{-1\pm \sqrt{1+12}}{2}$

$x=\frac{-1\pm \sqrt{13}}{2}$

so our two answers are:

a) $c=\frac{-1+\sqrt{13}}{2}$ and $c=\frac{-1-\sqrt{13}}{2}$