Question

According to LMVT, if a function f(x) is continuous on [a, b] and differentiable on the interval (a, b) then which of the following option should be correct for some value c from the interval (a,b)?( c can take any value from the interval (a,b) )

• A. $f^{\prime }\left(C\right)=\frac{f\left(a\right)-a}{f\left(b\right)-b}$
• B. $f^{\prime }\left(C\right)=\frac{f\left(a\right)-f\left(b\right)}{a-b}$
• C. $f^{\prime }\left(C\right)=\frac{f\left(b\right)-f\left(a\right)}{a-b}$
• D. None of the above

Answer 1

The correct option is B

$f^{\prime }\left(C\right)=\frac{f\left(a\right)-f\left(b\right)}{a-b}$

LMVT theorem states that if a function f(x) is continuous on [a, b] and differentiable on the interval (a, b) then we’ll have slope of the tangent drawn at some x = c where c ∈ (a, b) equal to the slope of secant joining points (a, f(a)) & (b, f(b)). Slope of tangent at x =c is f’(c). Slope of secant is the average rate of change of f(x) over the interval [a,b]

$\Rightarrow f^{\prime }\left(C\right)=\frac{f\left(a\right)-f\left(b\right)}{a-b}$