alleryThe function $f(x) = x^3 - 3x + 1$ has critical points at:
A$x = 0, 1$
B$x = \pm 1$
C$x = \pm \sqrt 3$
DNone
Answer & Solution
Correct answer: B. $x = \pm 1$
$f'(x) = 3x^2 - 3 = 0$ ⇒ $x = \pm 1$.
At $x = 1$ (local min, $f''(1) = 6 > 0$), at $x = -1$ (local max, $f''(-1) = -6 < 0$).
Keep practising Unit 4-5: Applications of Derivatives & Curve Sketching
Browse all Unit 4-5: Applications of Derivatives & Curve Sketching questions for AP Calculus AB, or explore more Mathematics topics.
Related questions
If the tangent to the curve $y = e^{2x}$ at point $(a, e^{2a})$ passes through origin, theIf $f(x) = x^x$ for $x > 0$, then $f$ has a minimum at $x = $:The function $f(x) = \dfrac{x}{1 + x^2}$ attains its maximum value on $[0, \infty)$ at:Find the **shortest distance** from origin to the curve $xy = 4$ (in the first quadrant):A rectangle is inscribed in a semicircle of radius $R$ with base on the diameter. The **ma$f(x) = 2x^3 - 3x^2 - 12x + 5$. The **local maximum value** occurs at:Find the approximate value of $ qrt{25.5}$ using differentials (given $ qrt{25} = 5$):If $f(x) = x^2 - 4x + 3$, the **interval** on which $f$ is decreasing is: