Solve for Variables to Find Continuity

Hypnos_16
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Homework Statement



y = 1 - 9x-2 / 1 - 3x-1 if x ≠ 3
y = a if x = 3

find the value of "a" that makes the graph Continuous at x = 3

Homework Equations


n/a


The Attempt at a Solution


I'm really not sure here, i think i must've missed this class or something, cause i just can't figure this out at all. I have two others to do too and i can't get any of them.
 
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You might want to compute

\lim_{x\rightarrow 3} \frac{1-\frac{9}{x^2}}{1-\frac{3}{x}}

and sketch the function around x=3 to see how it behaves.
 
Alright, i'll try that, is there also a way to solve it mathematically? Because i don't know what way he's looking for in the question.
 
What fzero suggests was "mathematical"! However:
The first thing I would do is multiply both numerator and denominator by x^2 to get
\frac{x^2- 9}{x^3- 3x}= \frac{(x- 3)(x+3)}{x(x-3)}
which is the same as \frac{x+ 3}{x} as long as x is not equal to 3. What is that when x= 3?
 
Taking the limit and interpreting it is mathematical. I suggested to sketch since it should help picture what you're doing.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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