Math Is Fun Forum

LuisRodg

Real Member

Registered: 2007-10-23

Posts: 322

Just took my Calc test --- got a couple questions.

Hello guys. Im just back home from taking a Calculus test and although I think I did ok, I missed some questions and wasn't sure about others.

1.) Integrate:

S((x^3+x-2)/(1+x^2))dx

I just didnt know what to do for that so I left it blank.

2.) f(x) = ax^2 +bx + c. Find a,b,c such that f(0)=4 is an absolute minimum of f in [-1,2]

I found that c=4 and b=0 but for the love of god I could not find the value of a....I did however find an inequality that showed that a>0 so I gave my answer as following:

b=0
c=4
a>0.

f(x) = ax^2 + 4, a>0.

That doesnt look like anything we've done in class but I think its better than leaving it blank. I did a couple of tests with different values of a and my answer held true for all values of a. Anyone care to expand on this?

3.) This was a Mean-Value-Theorem with f(x) = sqrt(x) -2x for x in [0,1]. I was able to do the problem completely and get the right answer etc. However, there was a part that asked me to show that f(x) meets all the requirements of the MVT. The requirements of the MVT are that f is continuous on [a,b] and differentiable on (a,b). I just said that f was a polynomial and therefore continuous and differentiable on [a,b] and (a,b) respectively. But I dont think thats right? I dont know, it seems I just pulled an easy answer off my math.

Anyways, I should be fine since I always get the highest grade and it curves up to a 100 so im sure I will still get an A but I just wanted to see the answers to this questions because when i cant do something it bugs me.

Thanks a lot.

TheDude

Member

Registered: 2007-10-23

Posts: 361

Re: Just took my Calc test --- got a couple questions.

1. Split this into 2 different integrals:

2. You are exactly right here.

3. It is not a polynomial, since square root is not a natural number exponent.  Split this into 2 functions: f(x) = g(x) - h(x), where g(x) = sqrt(x) and h(x) = 2x.  Showing that both g(x) and h(x) are differentiable (and, therefore, also continuous) on [0,1] shows that f(x) is too because it is a linear combination of g(x) and h(x).

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Wrap it in bacon