Question:medium
If $y = f(x)$ is continuous on $[0, 6]$, differentiable on $(0, 6)$, $f(0) = -2$ and $f(6) = 16$, then at some point between $x = 0$ and $x = 6$, $f'(x)$ must be equal to:
If $y = f(x)$ is continuous on $[0, 6]$, differentiable on $(0, 6)$, $f(0) = -2$ and $f(6) = 16$, then at some point between $x = 0$ and $x = 6$, $f'(x)$ must be equal to:
Show Hint
The Mean Value Theorem essentially says that the instantaneous slope (derivative) must equal the average slope (secant line) at some point in the interval.
Updated On: May 2, 2026
- $-18$
- $-3$
- $3$
- $14$
- $18$
Show Solution
The Correct Option is C
Solution and Explanation
Was this answer helpful?
0
Top Questions on Mean Value Theorem
- If \(f(x)\) and \(g(x)\) are differentiable functions for \(0 \leq x \leq 1\) such that, \(f(1)-f(0) = k(g(1)-g(0))\), \(k \neq 0\), and there exists a 'c' satisfying \(0<c<1\). Then, the value of \(\frac{f'(c)}{g'(c)}\) is equal to
- CUET (PG) - 2025
- Statistics
- Mean Value Theorem
- If Rolle's theorem holds for the function f(x) = x³ - ax² + bx - 4, x ∈ [1, 2] with f'(4/3) = 0, then ordered pair (a, b) is equal to :
- JEE Main - 2021
- Mathematics
- Mean Value Theorem
- Let A₁ and A₂ be two arithmetic means and G1, G2, G3 be three geometric means of two distinct positive numbers. Then $ G_{1}^{4} + G_{2}^{4}+ G_{3}^{4} +G_{1}^{2}G_{3}^{2} $ is equal to
- JEE Main - 2023
- Mathematics
- Mean Value Theorem
- Let the mean and the variance of 5 observations x1, x2, x3, x4, x5 be \(\frac{24}{5} \) and \(\frac{194}{25}\), respectively. If the mean and variance of the first 4 observations are \(\frac{7}{2}\) and a, respectively, then (4a + x5) is equal to
- JEE Main - 2022
- Mathematics
- Mean Value Theorem
- Want to practice more? Try solving extra ecology questions todayView All Questions
