Verify mean value theorem for the function f(x) = x2 + 3x + 1 in [-1,1 | Basic Mathematics 2081

Basic Mathematics20815 marksNumerical: mean value theoremMean value theorem

Verify mean value theorem for the function $f(x) = x^2 + 3x + 1$ in $[-1,1]$ . [5]

Verify mean value theorem for the function $f(x) = x^2 + 3x + 1$ in $[-1,1]$. [5]

Function: $f(x) = x^2 + 3x + 1$ - Interval: $[a, b] = [-1, 1]$ , so $a = -1$ , $b = 1$ If $f$ is continuous on $[a,b]$ and differentiable on $(a,b)$ , then there exists $c \in (a,b)$ such that: $f'(c) = \frac{f(b) - f(a)}{b - a}$ - $f(x) = x^2 + 3x + 1$ is a polynomial, hence continuous on

Function: $f(x) = x^2 + 3x + 1$ - Interval: $[a, b] = [-1, 1]$, so $a = -1$, $b = 1$ If $f$ is continuous on $[a,b]$ and differentiable on $(a,b)$, then there exists $c \in (a,b)$ such that: $$f'(c) = \frac{f(b) - f(a)}{b - a}$$ - $f(x) = x^2 + 3x + 1$ is a polynomial, hence continuous on

Verify mean value theorem for the function f(x) = x2 + 3x + 1 in [-1,1].