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### 67 Cards in this Set

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 definition of a critical value f'(c)=0 or if f'(c) does not exist than c is called a critical value of f Test for increasing and Decreasing Functions 1. a function f is increasing on an interval if f'(x) > 0 on that interval 2. A function f is decreasing on an interval if f'(x)<0 on that interval First Derivative Test= relative minimum C=critical value of a function f 1.If f'(x) changes from negative to positive at c, then f(c) is a relative minimum of f First Derivative test= relative maximum C= critical value of a function f if f'(c) changes from positive to negative at c, then f (c) is a relative maximum of f First derivative test= no relative max or min f'(x) does not change signs at c relative extrema relative extrema only occur at critical values but all critical values are not necessarily relative extrema Extreme Value Theorum If f is continuous on the closed interval [a,b], then f has botha minimum and a maximum on the interval Finding extrema on a closed interval 1. find values of f at the critical numbers of f in (a,b) 2. Find the values of f at the ENDPOINTS of the interval 3. the least of these values is the minimum, the greatst is the maximum Test for concavity=concave up if f''(x)>0 for all x in an open interval I, then the graph of f is concave upward on I Test for concavity= concave down if f''(x)<0 for all x in an open interval I, then the graph of f is concave downward on I definition of points of inflection a point P on a curve y=f(x) is called a point of inflection if f is continues on P and the curve changes from concave upward to concave downward or vice versa If (c, f(c)_ is a point of inflection of the graph of f, then either f''(x)=0 or f is not differentiable at x=c Second Derivative test= relative minimum f'(c)=0 and f''(x) exists on an open interval containing c 1. if f''(c)>0, the f(c) is a relative minimum Second derivative test= relative maximum f'(c)=0 and f''(x) exists on an open interval containing c if f''(c) <0, then f(c) is a relative maximum Second Derivative test=failed f'(c)=0 and f''(x) exists on an open interval containing c if f''(c)=0 the test fails use the first derivative test sketching a curve 1. domain 2. intercepts 3. symmetry 4. asymptote 5. intervals of increasing or decreasing 6. relative maximum and relative minimum 7. points of inflection and concavity domain what are the values of x for which f(x) is defined intercepts x intercept= let y=0 and solve for x y intercept= let x=0 and solve for y symmetry on the y axis if f(-x)=f(x) then f is an even function and the curve is symmetric about the Y AXIS symmetry about the origin if f(-x)=-f(x) then f is an odd function and the curve is symmetric about the origin symmetry around the x-axis when replacing y by -y yields an equivalent equation the curve is symetric about the x axis maximum or minimum value of a certain quantity is called... optimization optimization guidlines 1. draw a picture 2. assign a variable to the unknown quantity and write an equation for the quantity that is to be maximized or minimized 3. Use the given information to find a relationship between the variables. use these equations to eliminate all but one variable in the equation 4. use the first and second derivative tests to find critical points tangent line approximation line tangent to a curve at a point is the line that best approximates the curve near that point tangent line at point (c, f(c)) as approximation to the curve y=f(x) when x is near c tangent line approximation equation y-f(c)=f'(c)*(x-c) or y=f(c)+f'(c)*(x-c) derivative of √u 1/(2√(u))du/dx chain rule derivative of 1/u (-1/u^2)du/dx chain rule derivative of lnu 1/u du/dx chain rule derivative of logₐu (1/lna)*(1/u)*du/dx chain rule derivative of e^u e^u*du/dx chain rule the derivative of 7^u (ln7)*7^u*du/dx chain rule the derivative of sin(u) cos(u) the derivative of cos(u) -sin(u) the derivative of tan(u) sec²u the derivative of cot(u) -csc²u the derivative of sec(u) sec(u)tan(u) the derivative of csc(u) -csc(u)cot(u) the derivative of arcsin(u) 1/√(1-u²) the derivative of arccos(u) -1/(√(1-u²)) the derivative of arctan(u) 1/1+u² the derivative of arccot(u) -1/(1+u²) the derivative of arcsec(u) 1/(u*√(u²-1)) the derivative of arccsc(u) -1/(u*√(u²-1)) ∫u^n 1/(n+1)*u^(n+1)+C ∫1/u ln∣u∣+C ∫7^u (1/ln7)*a^u+C ∫tan(u) ln∣sec(u)∣+C ∫cot(u) ln∣sin(u)∣+C ∫sec(u) ln∣sec(u)+tan(u)∣+C ∫csc(u) ln∣csc(u)-cot(u)∣+C express f'(x) as a limit f'(x)= lim as h-->0 f(x+h)-f(x)/h express f'(a) as a limit f'(x)= lim as h-->0 f(a+h)-f(a)/h express f'(a) as limit approaches a f'(a)= lim as x--> a f(x)-f(a)/(x-a) because x=a+hand h=x-a when is a function differentiable a function f is differentiable at a only if f'(a) exisits. If a function is diferentiable at x=a, the f is continues at x=a, just because it is continuous does not mean it is differentiable instantaneous rate off change f'(x) when is f(x) not differentiable 1. when the graph has a corner when the graph is not continuous when the graph has a vertical tangent ln(xy) simplify lnx+lny lnx^p simplify plnx lnx/y simplify lnx-lny e^lnx simplify x slope of tangent line remember m=...? derivative tangent line is horizontal w/ respect to dy/dx dy/dx=0 tangent line is vertical w/ respect to dy/dx when denominator in expression for dy/dx is 0 f^(-1)'(x) 1/(f'(f^-1(x)) e^1 e ln(e) 1 -sin(arcsinx) -x Guidlines forsolving related rate problems 1. read problem carfully--draw diagram if possible. 2. Name the variables and constants 3. Write an equatioon that relates thevariables whose rates of change are give. Write anequation whoserates ofchange are to be determined. 4. combine two or more equations to get a single variable 5. usechainrule to differentiate bothsides with respect to t 6.substittute the given numerical information into the resulting equation and solve for the unknownrate