Problem Solving

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Problem Solving
1.a). Coordinates of points of inflection
f(x)=e-x22f'(x)=(-x)(e-x22)Second derivative = e-x22+e-x22.x2it follows that when 0=e-x22(x2-1)x=±1f(1)=1√eThe points of inflections are (1, 1/√e) and (-1, 1/√e)
b).f(x)=x+2sinxfirst derivative
=1+2cos x Second derivative
=2sinxGetting the local maxima and minima we equate the first derivative to zero
1+2cos x=0cos x=-12x=2π3, 4π3
evaluating the second derivative at x=2π3 and x=4π3-2sin2π3 =-2sin32=-√3.
Therefore x=2π3 is a maximum value
-2sin4π3 =-2sin-32=√3And x=4π3 minimum
C. f(x)=x13(x+4)The first derivative
=x13ddx(x+4)+x+4ddxx13=x13+(x+4)(13×23)=x13+(x3x23+43×23)f'(x)=43×23+4x133Using the product rule to get the second derivative
=43ddxx-23+x13+x-23+x13ddx43=43(-23x-53+13x-23)f”x=-89×53+49x23d. The intervals on which the function increases or decreases and local minimum
when the first derivative is equals to zero f’ , 43×23+4×133=04x133x-1 +1=0 it follow that x =0 or-1Function decreases for x<-1 and -1<x<0Increases for x>0Local minimum x=0e. concave down or up for the function and sketch a graph
f”x=-89×53+49×23 when f”x=0 therefore-492x-53-x-23=0 x=2by choosing x=0 and x=3 it follows that
f”0=43>0 hence -∞,2 the curve is concave upward )
And f”3=integer hence 2, ∞ the curve is concave upward
f. approximate intervals.
i. increasing. x>1ii. Decreasing x<1iii. concave up x=1iv. Concave down .does not concave downwards
2 a).

Squares to obtain a box with the maximum possible volume.
Let the size of square cut be x then the dimension of the box becomes
16-2x by 30-2x
volume=x(16-2x)( 30-2x)
Expand f(x)=4×3-92×2+480xFirst derivative f'(x)=12×2-184x+480Maximum and minimum derivative occurs when first derivative equals to zero
=(x-12)(x-103) hence x=103 or12We rule out x=12 since we cannot cut 12 inches twice from 16 inches of the square.
Hence our maximum possible volume is obtained when x=103b). Function fx=x4 -8×3 +6×2 +40x+3(i) local and absolute maximum and minimum
f’x=4×3 -24×2 +12x+40=(4x-20)(x+1)(x-2)x=5 or-1 or 2 When x=5, f5=-22, f4=3 , f6=27 hence x=5 is the local minimaWhen x=2, f2=59, f3=42 , f1=42 hence x=0 is the local maximaFor f-1=-22, f0=3 , f-2=27 hence x=0 is the local minimaii). X-coordinates at the points of inflections
f”x=12×2-48x+12Using the quadratic formula –b±b2-4ac2a x=2±23 which are the inflection points x

c. vertical asymptotes of f (x) =cotx vertical asymptote causes the denominator to be zero
cotx=cos xsin xdenominator is zero when x=0 or πThe equation is
x=kπ for some k ∈integers 2 (d). Equations for the horizontal asymptotes to the graph of y =2 tan-1 x
Solution.
tan-1x = arctan x at the interval (-π,2π,2)
therefore the horizontal asymptote for 2 arctan x is
y = ±π for y=2tan-1x(e) Asymptotes of f(x) =9×2+59×4+1dropping dominant terms we drop everything except big exponent
f(x) =9x29x4=9x23x2=3Hence the horizontal asymptote equation is y=33). Consider f(x)=x2-2x+4x-2a). the domain.
x<2b). the coordinates of the x-intercepts
if y=0, x is not defined hence no solutions c). coordinates of the y-intercepts
0,-2d).the coordinates of the critical points
(0,-2) and (4, 6)
e). the intervals on which the function increases or decreases
get the first derivative using the quotient rule (Chen, Fox & Lyndon 82)
f’x=x-2ddx x2-2x+4-[x2-2x+4ddxx-2]/(x-2)2first derivative=x2-4x(x-2)2 Where if
fx=0, the critical points are 0 and 4Function increases for x<0 , x>4 Decreases for 0<x<4f). the local maxima and the local minima
from the first derivative x=0 or 4When x=0, f0=-2, f-1=-73, f1=-3 , hence x=0 is the local maximaWhen x=4, f3=7 , f4=6, f-5=193hence x=4 is the local minima
g) The intervals for concave up and down
we get the second derivative using the quotient rule
f”x=8(x-2)3When second derivative=0 , x=2 by choosing two auxiliary points x=0 and x=3 it follows that
f”0=-1 hence -∞,2 the curve is concave downward )
And f”3=8 hence 2, ∞ the curve is concave upward i). Find limx→02x-sinxsinx L’Hospital Rule
limx→02-cosxcosx and cos 0 =1=(2-1)/1=1j). limx→∞1-1xx (exact value )using logarithm and exponential function
1-1xx=eln1-1xxTherefore limx→∞1-1xx=limx→∞eln1-1xx=e-1=1e The limit is 1e
Works Cited
Chen, K. T., Fox, R. H., & Lyndon, R. C. (1958). Free differential calculus, IV. The quotient groups of the lower central series. Annals of Mathematics, 81-95.