Suppose $f$ is continuous on $[a,b]$ and differentiable on $(a,b)$. If $f'(x)>0$ for every $x\in(a,b)$, then $f$ is
Astrictly increasing on $[a,b]$
Bstrictly decreasing on $[a,b]$
Cconstant on $[a,b]$
Dnot necessarily monotonic
Answer & Solution
Correct answer: A. strictly increasing on $[a,b]$
A positive derivative throughout an interval implies that the function increases as $x$ increases. Therefore $f$ is strictly increasing on the interval. A negative derivative would imply strict decrease.
Related questions
The angle made by the tangent of the curve $x = a$ ($t + in t$ cost); $y = a(1 + in t)^2A value of $C$ for which the conclusion of mean value theorem holds for the function $f(x)The value of $f(0)$ so that the function $f(x) = \frac{ qrt{1 + x} - qrt[3]{1 + x}}{x}$ b$\int_{-2}^{2}|1 - x^2|dx =$$\lim_{n\to \infty} um_{r = 1}^{n}1 - \frac{1}{n} e^{r}$ isLet $p(x)$ be a function defined on $R$ such that $p'(x) = p'(1 - x)$, for all $x \in [0, If $\frac{dv}{dx} = y + 3 > 0$ and $y(0) = 2$, then $y(\ln 2)$ is equal toThe equation $y^{2} \mathrm{e}^{xy} = 9 \mathrm{e}^{-3} \cdot x^{2}$ defines $y$ as a diff