The function is continuous on [-1,2] and differentiable on (-1,2) so the function satisfies the hypothesis for the mean value theorem.

The mean value theorem (MVT) states that there exists at least one c in [-1,2] such that `f'(c)=(f(2)-f(-1))/(2-(-1))` .

Now `(f(2)-f(-1))/(2-(-1))=(6-(-6))/3=4`

`f'(x)=3x^2+1` Setting f'(x)=4 we get:

`3x^2+1=4`

`3x^2=3`

`x^2=1==>x=+-1`

So the values of x where the slope of the tangent line to the function is equal to the average rate of change of the function over this interval are **-1 and 1.**

The graph of the function, the secant line, and the tangent lines:

Note that one of the tangent lines is the secant line.

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