First Dertivative TestMath165: Business CalculusRoy M. LowmanSpring 2010, Week6 Lec1Roy M. Lowman First Dertivative Testfirst derivative testanother application of f0Assume f(x) is not a straight line and at some x, f0(x) = 0, xis a critical number.What kind of point is the critical point (x, f(x))?There are four possibilities:Relative MinimumRelative Maximumone of two types of inflection points.Roy M. Lowman First Dertivative Testfirst derivative testanother application of f0Assume f(x) is not a straight line and at some x, f0(x) = 0, xis a critical number.What kind of point is the critical point (x, f(x))?There are four possibilities:Relative MinimumRelative Maximumone of two types of inflection points.Roy M. Lowman First Dertivative Testfirst derivative testanother application of f0Assume f(x) is not a straight line and at some x, f0(x) = 0, xis a critical number.What kind of point is the critical point (x, f(x))?There are four possibilities:Relative MinimumRelative Maximumone of two types of inflection points.Roy M. Lowman First Dertivative Testfirst derivative testanother application of f0Assume f(x) is not a straight line and at some x, f0(x) = 0, xis a critical number.What kind of point is the critical point (x, f(x))?There are four possibilities:Relative MinimumRelative Maximumone of two types of inflection points.Roy M. Lowman First Dertivative Testfirst derivative testanother application of f0Assume f(x) is not a straight line and at some x, f0(x) = 0, xis a critical number.What kind of point is the critical point (x, f(x))?There are four possibilities:Relative MinimumRelative Maximumone of two types of inflection points.Roy M. Lowman First Dertivative Testfirst derivative testanother application of f0Assume f(x) is not a straight line and at some x, f0(x) = 0, xis a critical number.What kind of point is the critical point (x, f(x))?There are four possibilities:Relative MinimumRelative Maximumone of two types of inflection points.Roy M. Lowman First Dertivative Testfirst derivative testf0= 0 ⇒ four posibilities-100102030405060-10 -5 0 5 10f1-30-20-1001020-10 -5 0 5 10f2-30-20-100102030-4 -3 -2 -1 0 1 2 3 4f3-30-20-100102030-4 -3 -2 -1 0 1 2 3 4f4Roy M. Lowman First Dertivative Testfirst derivative testf0= 0 ⇒ four posibilitiesThe slope pattern across the critical point can be used todetermine what kind of CP: Rel Min, REl Max or IP.This process is called The First Derivative Test.Slope pattern: −, 0, + ⇒ Relative Minimum.Slope pattern: +, 0, − ⇒ Relative Maximum.Slope pattern: +, 0, + ⇒ Inflection Point.Slope pattern: −, 0, − ⇒ Inflection Point.It is better to determine the shape of the graph using the signsof the slopes instead of trying to memorize the sign patterns.Roy M. Lowman First Dertivative Testfirst derivative testf0= 0 ⇒ four posibilitiesThe slope pattern across the critical point can be used todetermine what kind of CP: Rel Min, REl Max or IP.This process is called The First Derivative Test.Slope pattern: −, 0, + ⇒ Relative Minimum.Slope pattern: +, 0, − ⇒ Relative Maximum.Slope pattern: +, 0, + ⇒ Inflection Point.Slope pattern: −, 0, − ⇒ Inflection Point.It is better to determine the shape of the graph using the signsof the slopes instead of trying to memorize the sign patterns.Roy M. Lowman First Dertivative Testfirst derivative testf0= 0 ⇒ four posibilitiesThe slope pattern across the critical point can be used todetermine what kind of CP: Rel Min, REl Max or IP.This process is called The First Derivative Test.Slope pattern: −, 0, + ⇒ Relative Minimum.Slope pattern: +, 0, − ⇒ Relative Maximum.Slope pattern: +, 0, + ⇒ Inflection Point.Slope pattern: −, 0, − ⇒ Inflection Point.It is better to determine the shape of the graph using the signsof the slopes instead of trying to memorize the sign patterns.Roy M. Lowman First Dertivative Testfirst derivative testf0= 0 ⇒ four posibilitiesThe slope pattern across the critical point can be used todetermine what kind of CP: Rel Min, REl Max or IP.This process is called The First Derivative Test.Slope pattern: −, 0, + ⇒ Relative Minimum.Slope pattern: +, 0, − ⇒ Relative Maximum.Slope pattern: +, 0, + ⇒ Inflection Point.Slope pattern: −, 0, − ⇒ Inflection Point.It is better to determine the shape of the graph using the signsof the slopes instead of trying to memorize the sign patterns.Roy M. Lowman First Dertivative Testfirst derivative testf0= 0 ⇒ four posibilitiesThe slope pattern across the critical point can be used todetermine what kind of CP: Rel Min, REl Max or IP.This process is called The First Derivative Test.Slope pattern: −, 0, + ⇒ Relative Minimum.Slope pattern: +, 0, − ⇒ Relative Maximum.Slope pattern: +, 0, + ⇒ Inflection Point.Slope pattern: −, 0, − ⇒ Inflection Point.It is better to determine the shape of the graph using the signsof the slopes instead of trying to memorize the sign patterns.Roy M. Lowman First Dertivative Testfirst derivative testf0= 0 ⇒ four posibilitiesThe slope pattern across the critical point can be used todetermine what kind of CP: Rel Min, REl Max or IP.This process is called The First Derivative Test.Slope pattern: −, 0, + ⇒ Relative Minimum.Slope pattern: +, 0, − ⇒ Relative Maximum.Slope pattern: +, 0, + ⇒ Inflection Point.Slope pattern: −, 0, − ⇒ Inflection Point.It is better to determine the shape of the graph using the signsof the slopes instead of trying to memorize the sign patterns.Roy M. Lowman First Dertivative Testfirst derivative testf0= 0 ⇒ four posibilitiesThe slope pattern across the critical point can be used todetermine what kind of CP: Rel Min, REl Max or IP.This process is called The First Derivative Test.Slope pattern: −, 0, + ⇒ Relative Minimum.Slope pattern: +, 0, − ⇒ Relative Maximum.Slope pattern: +, 0, + ⇒ Inflection Point.Slope pattern: −, 0, − ⇒ Inflection Point.It is better to determine the shape of the graph using the signsof the slopes instead of trying to memorize the sign patterns.Roy M. Lowman First Dertivative Testfirst derivative testexampleTypical exam problem:f(x) = (x − 1)3+ 1 Find the location of any critical points anduse the first derivative test to determine what kind critial points.Find all critical points: set f0= 0 and solve for all x = xcf0(x) =ddx[(x − 1)3+ 1] =ddx(x − 1)3+ddx1 =3(x − 1)3−1ddx(x − 1) + 0 = 3(x − 1)2f0(x) = 3(x − 1)2= 0 Gives xc= 1 as the only criticalnumber.Now use the