Question

How do I find the numbers $c$ that satisfies the Mean Value Theorem for
$f(x) = {x^3} + x - 1$ on the interval $[0,3]$ ?

Answer 1

Hint : The mean value states that there exists a $c$in $(a,b)$ such that
$f'(c) = \dfrac{{f(a) - f(b)}}{{b - a}}$
Here in this case we write that
There exists a $c$ in $(0,3)$ such that
$f'(c) = \dfrac{{f(3) - f(0)}}{{3 - 0}}$
The answer can then be found by differentiating the given function and then substituting the value of $x$with $c$and then we easily put the values in the above equation along with the value of the given function at the boundary conditions which will finally help us to find the value of $c$. The differentiation can be done easily by using the standard formulae for differentiation

Complete step-by-step answer :
The mean value theorem in this case states that there exists a $c$ in $(0,3)$ such that

$$f'(c) = \dfrac{{f(3) - f(0)}}{{3 - 0}} \\\ f(3) - f(0)3 - 0 = \dfrac{{29 - ( - 1)}}{3} = 10 \;$$

Now to solve it further we first differentiate the given function so differentiating the given function at $c$ we get
$f'(c) = 3{c^2} + 1$
Now substitute the value of $f'(c)$ in the equation of mean value theorem we get
$3{c^2} + 1 = \dfrac{{f(3) - f(0)}}{{3 - 0}}$
Now we put the value of boundary conditions upon the above equation we get
$3{c^2} + 1 = \dfrac{{29 - ( - 1)}}{3}$
Which upon solving gives,
$3{c^2} + 1 = 10$
We get $3{c^2} = 9$
Which gives us the value of $c$ as
$c = \pm \sqrt 3$
But since $- \sqrt 3$does not fall in our given interval of $\left( {0,3} \right)$we ignore this value of $c$ that we calculated and thus the only remaining value of the $c$is $\sqrt 3$. So the value of $c$ that satisfies the mean value theorem of the given function is $\sqrt 3$.
So, the correct answer is “$\sqrt 3$”.

Note : Remember to check if some value of $c$does not fall in the given interval then we just ignore that value and take only that value which lies inside the given boundary conditions.