Using the Mean Value Theorem, find the values of $x_0\in [a,b]$ when
$f(x)=x^k$ and $k=1,2$ and $3 \;\;\;(**).$
I want to use the Mean Value Theorem to solve (**). Please, can anyone help solve this? I'm so sorry for the wrong information I gave earlier!
The MVT states $$ f(b)-f(a)=f'(x_0)(b-a) $$ for some $x_0$ between $a$ and $b$. Just plug in $f$ for a given $a$ and $b$, and find out what $x_0$ has to be.
Here's the calculation for $k=2$; I'll leave $k=1$ and $k=3$ to you. When $k=2$, we're talking about $f(x)=x^2$, so $f'(x)=2x$. Plug in to MVT:
$$b^2-a^2=f(b)-f(a)=f'(x_0)(b-a)=2x_0(b-a).\tag1$$ Factor the LHS of (1) as $(b-a)(b+a)$, then divide the outsides of (1) through by $b-a$: $$ b+a=2x_0 $$ to conclude $x_0=(b+a)/2$. Interpretation: On the graph of $f(x):=x^2$, the secant line connecting $(x,y)=(a,a^2)$ to $(b, b^2)$ has the same slope as the tangent to $f$ at $x_0=(a+b)/2$. Normally it isn't possible to determine the exact value of $x_0$ that is guaranteed to exist by the MVT, but it's possible when $f$ has this simple form.