Author Tutor CourseDate 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)(13x23)=x13+(x3x23+43x23)f'(x)=43x23+4x133Using the product rule to get the second derivative =43ddxx-23+x13+x-23+x13ddx43=43(-23x-53+13x-23)f''x=-89x53+49x23d. The intervals on which the function increases or decreases and local minimum when the first derivative is equals to zero f' , 43x23+4x133=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=-89x53+49x23 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

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