Math final index card - If graph f(x is bounded by m&M...

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If k is a differentiable function and is always concave up, then k (a) ≤ (k(b) − k(a))/(b-a) whenever a < b. Linear approx/linearization: g(x)=g(a) + g (a)(x−a) The critical points of H are the values of t for which H is zero or undefined. Now, H (t)=0 when t=3 or t=−450, and H (t) is undefined for t=25. Now we need to check whether each of these is a local max, local min, or neither. These three critical points split the domain into four intervals: (∞, −450), (−450, 3), (3, 25), and (25,∞). We will find out the sign of H for each interval by checking the sign of each factor- Then table- then results. g ′′ (x) ≈ ∆g (x)/(∆x) C(t)dt is the average temperature, in degrees Fahrenheit, of the can of soda during the first minute it is in the refrigerator. (Integral 1-0) Radians/time is the unit for degrees Continuous exponential is F=Qe^(kt) k=rate or F=QA^t A=+ or - rate If given average cost function of a(q) multiply function by q to get C(q)
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Unformatted text preview: If graph f(x) is bounded by m&M m ≤ f(x) ≤ M (def integral lies between m(b-a) and M(b-a) ∆ t=(b-a)/n F Inc Dec CU CD Linear Horiz F’ +-Inc Dec Horiz X-axis (0) If a=b, then the integral will be zero since in that case, there will be no area under the curve. If I run 5 miles to the park and 5 miles home, then the total distance traveled is 10 miles (= total area under the curve) but my distance from my starting point is 0 miles(= integral) Step 1- Draw a pic & label at most two variables Step 2- Express the quantity you want to maximize or minimize in terms of your variables V=3x^2h Step 3- Constraint 10=3x^2+8 h x. Solve for one variable in terms of the other Step 4- Substitute- DON’T PLUG IN UNTIL HAVE THE FINAL FORMULA Find distance accurate within 20 ft: L − R = (v(0) − v(8))∆t = 80∆t 80∆t=20 ∆t=.25 sec...
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Math final index card - If graph f(x is bounded by m&M...

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