Let $f:\mathbb{R}\to\mathbb{R}$ be a twice differentiable function (meaning $f'$ and $f''$ exist) such that $f(0)=f'(0)=0$, $f(1) = 1$, and $f'(1) = 4$.

Each of the following statement follows from either the Intermediate Value Theorem, the Extreme Value Theorem, or the Mean Value Theorem. Determine which one in each case.

$f''(x)=f'(x)+3$ for some $x$ in $[0,1]$ ---------EVT

$f$ and $f'$ are bounded on $[0,1]$--------------MVT

$f''(x) = 4$ for some $x$ in $[0,1]$.-------MVT

$f'(x) = 2$ for some $x$ in $[0,1]$.=----------------IVT

I am not centain about my answers, escpecially the first one, I am trying to understand these theorems and please let me know if I am correct! Thanks! I appreciate your help!


First question: Let $g(x)=f'(x)-f(x)$. Then $g(0)=0$ and $g(1)=3$. So by MVT there is a $c$ between $0$ and $1$ such that $g'(c)=\frac{g(1)-g(0)}{1-0}=3$.

That says that $g'(c)=3$, that is, that $f''(c)-f'(c)=3$.

Second question: The boundedness of $f$ and $f'$ follow from the EVT. Since the second derivative exists, $f'$ is continuous. Also, $f$ is continuous. So each attains a max and a min on our interval, and as a consequence each is bounded.


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