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(MTH 250). Calculus. Lecture 13. Previous Lecture’s Summary. Critical points Increasing & decreasing Functions Concavity & inflection points Strategy of Graphing. Today’s Lecture. Recalls: Relative Extrema Extreme Value Theorem First Derivative Test Second Derivative test
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(MTH 250) Calculus Lecture 13
Previous Lecture’s Summary • Critical points • Increasing & decreasing Functions • Concavity & inflection points • Strategy of Graphing
Today’s Lecture • Recalls: • Relative Extrema • Extreme Value Theorem • First Derivative Test • Second Derivative test • Rolle’s Theorem • Mean Value Theorem
Recalls Theorem: Let be a function that is continuous on a closed interval and differentiable on the open interval • If for every value of in then is increasing on • If for every value of in , then is decreasing on . • If for every value of in , thenis constant on . • The points where are called critical points.
Recalls • Theorem: (Test for convavity) Let be a function whose second derivative exists on an open interval • If for all on , then the graph of is concave up on • If for all on , then the graph of is concave down on • The points where changes its sign are called inflection points. The points where or undefined are possible candidate for inflection points.
Relative extrema We say a continuous function has • a relative (local) maximum at if for near, • a relative (local) minimum at if for near, • a relative extremum at if it has either a relative maximum or a relative minimum at . A relative/local max/min may not be a global max/min.
Relative extrema Examples: • has a relative maximum at and a relative minimum at . • has relative minima at and and a relative maximum at .
Extreme value theorem Theorem: If is continuous on a closed interval , then attains an absolute maximum value Mand an absolute minimum value in that is, there are numbers and in with for every other in . Fermat's Theorem: If has local maximum or minimum at , and exists, then .
Extreme value theorem A function might have derivative equal to zero at a number c, and therefore the function will have a horizontal tangent line at c, but the function will not have a local maximum or minimum at that point.
Extreme value theorem Example: Find all critical points of . Solution: The function is continuous everywhere and its derivative is Thus are critical points and is a stationary point.
Extreme value theorem • Example Find the absolute maximum and minimum values of on the interval . • Solution • Evaluate the functionat the endpoints and critical numbers. • We can conclude that the absolute minimum is and the absolute maximum is
Extreme value theorem • Relative extremamust occur at critical points, • It does not meanthat a relative extremum occurs at every critical point. • For the critical point where we have relative extremum the derivatives have opposite signs on the two sides of , • For the critical points where we don’t‘ have relative extremum the signs of the derivatives have same sign on both sides. • A function has a relative extremum at those critical points where f changes sign ???
First derivative test • Theorem: Suppose that is continuous at a critical point . • If on an open interval extending left from and on an open interval extending right from , then has a relative maximum at . • If on an open interval extending left from and on an open interval extending right from , then has a relative minimum at . • If has the same sign on an open interval extending left from as it does on an open interval extending right from , then does not have a relative extremum at .
First derivative test Example: Find the dimensions of the rectangle with maximum area that can be inscribed in the ellipse . y 16x2+9y2=144 Solution: Let be the vertex of the rectangle in the first quadrant. So is in the interval Area of rectangle is Isolating from the ellipse So P(x,y) x
First derivative test Cont.. By first derivative test For there is a local maximum. But being this the only critical point in it has to be a absolute maximum. The dimension of the rectangle are + - 2 2.12 2.2
Second derivative test • Theorem: Suppose that is twice differentiableat a critical point i.e. • If , then has a relative maximum at . • If , then has a relative minimum at . • If then the test is inconclusive; i.e. may have a relative maximum, a relative minimum, or neitherat .
Second derivative test Example: Proof:
Second derivative test Cont. We check now the sign of first derivative before and after As does not change sign, is neither a relative maximum nor a relative minimum to the function
Second derivative test Example: A can (cylinder) with volume of has to be constructed. Find the dimension of the cheapest (using the least material) can. (Assume that the radius r has to satisfy the inequality ). Solution: We have to minimize the total surface of the Cylinder. Isolating h from we get . Replacing in formula we get . To find critical numbers, put . Solving So and Therefore the radius of the cheapest can is
Rolle’stheorem • Let be a function that satisfies the following two conditions: • is continuous on the closed interval • is differentiable on the open interval • Then there exists a number in such that • Remark: If either the maximum or the minimum occurs at a point between and , then , and we have found a point for Rolle’s theorem.
Rolle’stheorem • Proof: If in , then at every point in • If at some point and is continuous on , it follows from the extreme-value theorem that has an absolute maximum on . The maximum cannot occur at end points as Thus the absolute maximum at an interior point which is a critical point of , and since f is differentiable on , this critical point must be a stationary point, that is
Rolle’stheorem Example: Considere the polynomial function . It is continuous at every point and is differentiable on the open interval . Since , by Rolle’s theorem must be zero at least once in . In fact, is zero twice in this interval, once at and again at .
Rolle’stheorem The conditions of Rolle’s theorem are essential. If they fail at even one point, the graph may not have a horizontal tangent
Mean value theorem • Let be a function that satisfies the following two conditions: • is continuous on the closed interval • is differentiable on the open interval • Then there exists a number in such that
Mean value theorem Proof: Let .Then is continuous on , differentiable on and By Rolle’s theorem, there exists such that . .
Mean value theorem Example: Consider. The functioniscontinuous and differentiableeverywhere. In particulariscontinuous on and differentiable onMoreover, and By mean value theoremthereis a suchthat
Mean value theorem Example Suppose that and that for all value of (therefore is differentiable everywhere). How large can possibly be?
Mean value theorem Physical interpretation: If we think of the number as the average change in over and as an instantaneous change, then the Mean value theorem says that at some interior point the instantaneous change must equal the average change over the entire interval.
Mean value theorem Example: If a car is accelerating from zero takes sec to go its average velocity for the interval is , At some point during the acceleration, the Mean value theorem says, the speedometer must read exactly . As shown in the figure.
Lecture Summary • Relative Extremum • Extreme Value Theorem • First Derivative Test • Second Derivative test • Rolle’s Theorem • Mean Value Theorem