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How to Find the Domain of Any Function (NancyPi)

Duration: 12:40Views: 914.1KLikes: 23.2KDate Created: May, 2018

Channel: NancyPi

Category: Education

Tags: domain of a functionrulesalgebratrickhow to findexplanationall casesexampleproblemsolutionradicandfindfunctionhow do yousquare rootalgorithmall real numberslinearradicaltestprecalculusequationquadratic expressionnancyzerosfractiontutordenominatornumeratorset notationpolynomialdomainsolvealgebra 2rational functionreal numbersinfinitygeometrynancypiinterval notation

Description: MIT grad shows a surefire way to find the domain of any function. To skip ahead: 1) For POLYNOMIAL only, skip to time 0:45. 2) For FRACTION only, skip to time 1:44. 3) For SQUARE ROOT only, skip to 5:08. 4) For SQUARE ROOT ON BOTTOM of Fraction, skip to 8:44. 5) For SQUARE ROOT ON TOP of Fraction, skip to 10:09. Nancy formerly of MathBFF explains the steps. For how to SOLVE QUADRATIC equations by FACTORING, jump to: youtu.be/Z5MnP9da4EM For how to SOLVE INEQUALITIES, jump to: youtu.be/ONuIIvgkE3c Follow Nancy on Instagram! instagram.com/nancypi Twitter: twitter.com/nancypi For Domain, you just need to watch out for Square Roots and/or Fractions! If your function is... 1) a POLYNOMIAL (no square roots or fractions), i.e. f(x) = x^2 + 3x + 1, domain is "all real numbers." 2) a FRACTION (w/ no square root), i.e. f(x) = (2x+1)/(x^2+5x+6). Set bottom "not equal" to zero. Solve for x. 3) a SQUARE ROOT (w/ no fraction), i.e. f(x) = sqrt(x+1). Set part under root "greater than or equal to zero". 4) a FRACTION w/ A SQUARE ROOT ON BOTTOM, i.e. f(x) = (x^2+2x+3)/(sqrt(x+1)). Set part under root "greater than zero." Solve the inequality. NOTE THE DIFFERENCE: You must set it greater than zero, not "greater than OR EQUAL TO zero" because we can't have zero (or square root of zero) in the denominator. 5) a FRACTION w/ A SQUARE ROOT ON TOP, i.e. f(x) = (sqrt(x+1))/(x^2+4), i) Set the bottom "not equal to" zero & solve; ii) Set what's under the root "greater than or equal to zero" & solve; iii) Take intersection of your two solutions. For more of my math videos, check out: nancypi.com

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