Problem 1 - Cornell University



Math 118 - Sample Midterm

Problem 1.

f(x)=(9-x2)2.

Find the functions’ critical values; determine where the function is increasing and where it’s decreasing; find its inflection points; determine where it is concave up and where it is concave down; sketch the graph.

|f’(x)=3(9-x2)*-2x=-6x(9-x2) |f’’(x)=6[1*(9-x2)+x*-2x] = |

|f’(x)=0 => x=-3,0,3 are the critical pts. |=6[9-x2-2x2]=6*(9-3x2) |

|f’(-4) x x=±(3 – inflection points |

|f’(-2)>0 => -3 -3 is min & x x2=18/2=9 => x=±3 => x=3.

p’’(x) = 2+18/x2. p’’(3)=2+18/9=2+2=4 > 0 so min point.

p(1)=1-18ln(1)=1-18*0=1; p(20)=400-18ln(20)≈400-3=397.

So max profit is 397$ achieved on production of 20 lava lamps.

Problem 5.

Find the integrals of the following functions:

1. ( [ex-ln(x2)+((x2-2)/x½)] dx

2. ( [(12y-16)/(3y2-8y-5)] dy

3. ( [ue-u/2] du

1. ( [ex-ln(x2)+((x2-2)/x½)] dx = ( [ex-ln(x2)+(x2/x½)-(2/x½)] dx =

= ( ex dx - ( 2ln(x) dx + ((x1.5) dx - ((2*x-½)] dx =

= ex – 2( ln(x) dx + x2.5/2.5 – 2*x0.5/0.5 + C = ex – 2( ln(x) dx + x2.5/2.5 – 4x0.5 + C

Now do ( ln(x) dx by parts:

Set U=ln(x), dU=(1/x)dx and then dV is equal to the rest i.e. dV=dx. V=( dx=x

( ln(x) dx = xln(x) - ( x(1/x)dx = xln(x) - ( 1dx = xln(x) – x

And the whole integral becomes: ex – 2(xln(x)-x) + x2.5/2.5 – 4x0.5 + C.

2. ( [(12y-16)/(3y2-8y-5)] dy

Set x=(3y2-8y-5). dx=(6y-8)dy. dy=dx/(6y-8)

So ( [(12y-16)/(3y2-8y-5)] dy = ([(12y-16)/x]*[dx/(6y-8)] = ([(12y-16)/(6y-8)]*1/x* dx

= ((2/x)dx = 2((1/x)dx = 2ln(x) + C = 2ln(3y2-8y-5) + C.

3. ( [ue-u/2] du

Set U=u, dV= e-u/2du; V=(e-½ udu= -2e-u/2du ;

( [ue-u/2] du = (UdV = UV-(VdU = -2ue-u/2 - (-2e-u/2du = -2ue-u/2 +2(-2e-u/2) + C =

= -2ue-u/2 - 4e-u/2 + C = -2e-u/2(u-2) + C

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