- Thread starter Jkacson
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If you don't find a theorem that applies, look for a

The best way for us to help you would be if you make some attempt at answering each part and tell us your thinking - even if it's mostly a guess. Then we can suggest ideas you might be missing. (The exercise will be most beneficial to you if it makes you think a lot! That's the whole point, so we won't give away the answer before you've had a chance to learn from it!)

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Except for c), the others should be simple.View attachment 11721So for number 18 on here I'm not sure how to even figure out if they're true or not

a) Look here

c) Use the mean value theorem.

d) \(\displaystyle f(x)=x^3,~x=0\)

e) use chain rule.

b) false because f(x) and g(x) is not equal to (f*g)(x)

c) I looked over the mean value, but I'm still not understanding how to implement it

d) I got true because critical points occur at the first derivative equal to zero and critical points are where local extrema is

e) I ended getting false from applying the chain rule

I'm also not very good at calculus so I wouldn't be surprised if I'm still doing these wrong

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(a) You haven't described your counterexample very well, but you may be thinking about the right thing. Can you give a

b) false because f(x) and g(x) is not equal to (f*g)(x)

c) I looked over the mean value, but I'm still not understanding how to implement it

d) I got true because critical points occur at the first derivative equal to zero and critical points are where local extrema is

e) I ended getting false from applying the chain rule

I'm also not very good at calculus so I wouldn't be surprised if I'm still doing these wrong

(b) Again, give a

(c) Suppose that f(a) = f(b). What does the MVT tell you?

(d) Are you sure that

(e) Can you show the details of what you did?

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For (a) you do not need symmetry. Must a function as described in part (a) have an absolute maximum point? Can it have more than one? In my opinion instead of thinking about this you should just draw a continuous function from a to b (do not lift your pen) and see if it has an absolute maximum value. If it has more than one absolute maximum then you should know the answer to the question. If your graph has only one absolute max than ask yourself if you can modify it so it has two or more absolute extremes.

b) false because f(x) and g(x) is not equal to (f*g)(x)

c) I looked over the mean value, but I'm still not understanding how to implement it

d) I got true because critical points occur at the first derivative equal to zero and critical points are where local extrema is

e) I ended getting false from applying the chain rule

I'm also not very good at calculus so I wouldn't be surprised if I'm still doing these wrong

b) Are you saying that f(x) = g(x) = (f*g)(x) can never happen? What if f(x) = g(x) = x, then (f*g)(x) = x AND if f(x) = g(x) = 7, then (f*g)(x) = 7. But for your question move away from that. We know that g(a) is a rel max, say its 104. What if anything do we know about f(104)? Note that (f*g)(a) = f(g(a)) = f(104).

c) Personally I would not use the MVT. What does it mean if a\(\displaystyle \neq\)b, then f(a) \(\displaystyle \neq\) f(b), ie if a and b are different then f(a) and f(b) must be different? Think carefully about what does that mean (hint: does f(x) = x

d) Look at the graph of f(x) = x

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c) Suppose that \(\displaystyle a<b\) having the property that \(\displaystyle f(a)=f(b)\).View attachment 11721So for number 18 on here I'm not sure how to even figure out if they're true or not

By the mean value theorem, \(\displaystyle \exists c\in(a,b)\) having the property that \(\displaystyle f'(c)=\frac{f(b)-f(a)}{b-a}=0.\)

That is contrary to the given conditions on \(\displaystyle f\).

d) This has already been posted, but to reinforce it. Inflection points are critical points but by almost universal agreement they are not local extrema. That is true of \(\displaystyle f(x)=x^3\) at \(\displaystyle x=0\).

e) Given that \(\displaystyle f(-x)=f(x)\) (i.e. \(\displaystyle f\) is an even function) the it follows that:

\(\displaystyle \begin{align*}f(x)&=f(-x) \\f'(x)&=-f'(-x)\\-f'(x)&=f'(-x) \end{align*}\)

Which shows \(\displaystyle f'\) is an odd function.