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- lim y approaches 2 .............. y + 2/[y^2 + 5y + 6]Find the global minimum and maximum of the continuous function f(x) = x2 − 8 ln(x) on [1, 5]. (Round your answers to two decimal places.) global minimum value? global maximum value?box with a square base and open top must have a volume of 42592 cm3. We wish to find the dimensions of the box that minimize the amount of material used.First, find a formula for the surface area of the box in terms of only x, the length of one side of the square base.[Hint: use the volume formula to express the height of the box in terms of x.]Simplify your formula as much as possible.A(x)= Next, find the derivative, A'(x).A'(x)=Now, calculate when the derivative equals zero, that is, when A′(x)=0. [Hint: multiply both sides by x2.]A′(x)=0 when x=We next have to make sure that this value of x gives a minimum value for the surface area. Let's use the second derivative test. Find A"(x).A"(x)=Evaluate A"(x) at the x-value you gave above.
- A box with a square base and open top must have a volume of 171500 cm3cm3. We wish to find the dimensions of the box that minimize the amount of material used.First, find a formula for the surface area of the box in terms of only xx, the length of one side of the square base.[Hint: use the volume formula to express the height of the box in terms of xx.]Simplify your formula as much as possible.A(x)=A(x)= Next, find the derivative, A'(x)A′(x).A'(x)=A′(x)= Now, calculate when the derivative equals zero, that is, when A'(x)=0A′(x)=0. [Hint: multiply both sides by x2x2.]A'(x)=0A′(x)=0 when x=x=We next have to make sure that this value of xx gives a minimum value for the surface area. Let's use the second derivative test. Find A"(x)(x).A"(x)=(x)= Evaluate A"(x)(x) at the xx-value you gave above.NOTE: Since your last answer is positive, this means that the graph of A(x)A(x) is concave up around that value, so the zero of A'(x)A′(x) must indicate a local minimum for A(x)A(x). (Your boss is…Consider the revenue function R(x)=x^2 ln (x) a) find where the revenue is zero. b) find the quantity that yeilds the minimum revenue. c) find the point of diminishing returns.f(x) = 3√x−1. (a). Use f′(a) = limx→a f(x)−f(a)/x−a or f′(a) = lim h→0 f(a+h)−f(a)/ h to find f′(4).(b). Find the equation of the tangent line for the graph of f at x= 4.
- The cost C (in dollars) of producing x units of a product is C = 1.60x + 9,000. (a) Find the average cost function C. C = (b) Find C when x = 1,000 and when x = 10,000. C(1,000) =$ per unit C(10,000) =$ per unit (c) Determine the limit of the average cost function as x approaches infinity. lim x→∞ C(x) = Interpret the limit in the context of the problem. As more and more units are produced, the average cost per unit (in dollars) will approach $16) The cost of fighting crime in a country increased significantly during the period 1982–1999. Total spending on police and courts can be approximated by P(t)= 1.797t + 29.16 billion dollars (2 ≤ t ≤ 19) C(t)= 1.037t + 10.57 billion dollars (2 ≤ t ≤ 19), respectively, where t is time in years since 1980. Compute lim t→+∞ P(t)/C(t) to two decimal places. (If an answer does not exist, enter DNE.) _______________ Interpret the result. If the trend continues indefinitely, the annual spending on police will be ____________ times the annual spending on courts in the long run.Find all relative extrema of the function. (If an answer does not exist, enter DNE.) f(x) = 5x - 60x^1/3 Relative Maximum (x,y) =? Relative Minimum (x,y) =?
- (a) Find the slope of the curve y=x^2−4x−5 at the point P(3,−8) by finding the limit of the secant slopes through point P. (b) Find an equation of the tangent line to the curve at P(3,−8).The air in a room with volume 180 m3 contains 0.15% carbon dioxide initially. Fresher air with only 0.05% carbon dioxide flows into the room at a rate of 2 m3/min and the mixed air flows out at the same rate.Find the percentage of carbon dioxide in the room as a function of time t (in minutes).p(t) = What happens with the percentage of carbon dioxide in the room in the long run? lim t → ∞ p(t) = %Linearizations at inflection points Show that if the graph of a twice-dierentiable function ƒ(x) has an inflection point at x = a, then the linearization of ƒ at x = a is also the quadratic approximation of ƒ at x = a. This explains why tangent lines fit so well at inflection points.