what values of c and d make f(x) continuous everywhere?

Question

Given f(x)= (dx^2+6cx)/x if x<0, dx^2+cx+5 if 0≤x≤2, dx-c if x>2.

What values of c and d make f(x) continuous everywhere?

Answer 1

f(x)=(dx²+6cx)/x=dx+6c when x<0.

When x=0, f(x)=5, and f(2)=4d+2c+5.

As x→0^- (the left of zero), f(x)→6c, so for continuity 6c=5, c=⅚.

Therefore f(2)=4d+5/3+5.

When x→2⁺ (the right of 2), f(x)→2d-⅚.

For continuity, 2d-⅚=4d+5/3+5, -15/2=2d, d=-15/4.

So f(x)=

{ -15x/4+5, | x<0

{ -15x²/4+5x/6+5 | 0≤x≤2

{ -15x/4-⅚ | x>2

So what does this graph look like? All negative values of x lie on a straight line which reaches the y-axis at y=5. Points then take a ride down the right arm of an inverted parabola, crossing the x-axis and plummeting downwards until the parabola meets a line parallel to the first line at x=-25/3. Points continue indefinitely along this linear track.