A function is differentiable at x = a if: — f'(a) exists (the derivative limit exists).Differentiability requires the derivative limit to exist. Differentiability implies continuity, but not vice versa.
If f is differentiable at x = a, then f is: — Necessarily continuous at a.Differentiability implies continuity. A differentiable function cannot have jumps or holes.
At a corner (like |x| at x=0), the function: — Is not differentiable.At a corner, the left and right derivatives differ, so the limit doesn't exist — the function is not differentiable at that point.
The equation of the tangent line to y = x² at (2, 4) is: — y = 4x − 4.Slope = 4 at x=2. Using point-slope: y − 4 = 4(x − 2) → y = 4x − 4.
The difference quotient for f(x) = x² at x = 1 with h = 0.1 is approximately: — 2.1.[(1.1)² − 1²]/0.1 = [1.21 − 1]/0.1 = 0.21/0.1 = 2.1. As h→0 this approaches f'(1) = 2.
The instantaneous velocity of an object at time t equals: — The derivative of the position function.Instantaneous velocity v(t) = s'(t) — the derivative of the position function with respect to time.
f'(c) = 0 means: — The tangent line at c is horizontal.f'(c) = 0 means the slope of the tangent is zero — the tangent line is horizontal (potential max, min, or inflection point).
The average rate of change of f(x) = x² over [1, 3]: — 4.[f(3)−f(1)]/(3−1) = [9−1]/2 = 8/2 = 4.
f''(c) > 0 at a critical point implies: — Local minimum.Second Derivative Test: f'(c)=0 and f''(c)>0 implies local minimum (concave up). f''(c)<0 implies local maximum.
Quotient rule: d/dx[f/g] = ? — (f'g − fg')/g².Quotient rule: (f'g − fg')/g². Remember: "low d-high minus high d-low, over the square of what's below."
Chain rule: d/dx[f(g(x))] = ? — f'(g(x))·g'(x).Chain rule: f'(g(x))·g'(x). Differentiate the outer function (at the inner), multiply by the derivative of the inner.
Why can't we simply solve for y first in all implicit equations? — Both B and C.Some equations (like x³+y³+xy=1) cannot be solved for y in closed form. Also, implicit differentiation is often faster even when explicit form exists.
Implicit differentiation is based on the: — Chain rule.The chain rule: when differentiating y (a function of x) with respect to x, we get dy/dx. This is the essence of implicit differentiation.
Find where the tangent to x² + xy + y² = 7 is horizontal: — Where dy/dx = 0.Horizontal tangent occurs where dy/dx = 0 (numerator of dy/dx = 0, denominator ≠ 0).
For x³ + y = 4, the slope at (1, 3) is: — −3.3x² + dy/dx = 0 → dy/dx = −3x². At (1,3): dy/dx = −3(1)² = −3.
Implicit differentiation works even when the function is defined by a relation that is: — Both A and B.Implicit differentiation works for any differentiable relation, even curves that fail the vertical line test like circles.
For y² = x, dy/dx = ? — 1/(2y).2y·(dy/dx) = 1 → dy/dx = 1/(2y). This is consistent with the explicit form: y=√x → y'=1/(2√x).
The tangent to the unit circle x²+y²=1 at (√2/2, √2/2) has slope: — −1.dy/dx = −x/y = −(√2/2)/(√2/2) = −1.
A circle's radius grows at 2 cm/s. dA/dt when r = 5: — 20π cm²/s.A = πr². dA/dt = 2πr(dr/dt) = 2π(5)(2) = 20π cm²/s.
A cube's side shrinks at 1 m/s. How fast is the volume decreasing when s = 3 m? — −27 m³/s.V = s³. dV/dt = 3s²(ds/dt) = 3(9)(−1) = −27 m³/s. Negative since shrinking.
For the balloon: V = (4/3)πr³ and dV/dt = 100 cm³/s. Which equation gives dr/dt? — Both are equivalent.dV/dt = 4πr²(dr/dt) → dr/dt = dV/dt/(4πr²) = 100/(4πr²). Options A and B are the same.
A point moves along y = x². At x = 3 and dx/dt = 2, find dy/dt: — 12.y = x². dy/dt = 2x(dx/dt) = 2(3)(2) = 12.
For x² + y² = r² (constant), differentiating with respect to t gives: — 2x(dx/dt) + 2y(dy/dt) = 0.Differentiating implicitly with respect to t: 2x(dx/dt) + 2y(dy/dt) = 0.
If volume V of a sphere increases at 36π cm³/s, find dr/dt when r = 3: — 1 cm/s.dV/dt = 4πr²(dr/dt). 36π = 4π(9)(dr/dt) → dr/dt = 36π/36π = 1 cm/s.
If A = (1/2)bh and b is constant, dA/dt = ? — (1/2)b(dh/dt).If b is constant, d/dt[bh] = b(dh/dt). So dA/dt = (1/2)b(dh/dt).
A particle moves along y = 2x + 1. If dx/dt = 3, find dy/dt: — 6.y = 2x + 1. dy/dt = 2(dx/dt) = 2(3) = 6.
The perimeter of a square grows at 4 cm/s. When the side is 10 cm, how fast is the area growing? — 20 cm²/s.P = 4s, so dP/dt = 4(ds/dt) = 4 → ds/dt = 1 cm/s. A = s², dA/dt = 2s(ds/dt) = 2(10)(1) = 20 cm²/s.
The Second Derivative Test: at a critical point c with f'(c) = 0, if f''(c) < 0, then c is: — A local maximum.f''(c) < 0 means the function is concave down at c — like the top of a hill — so c is a local maximum.
An inflection point occurs where: — f''(x) changes sign.An inflection point is where the concavity changes — f''(x) changes sign (positive to negative or vice versa).
If f''(x) > 0 on an interval, f is: — Concave up.Positive second derivative ↔ concave up (bowl shape).
To find the tangent line to y = x³ at (2, 8): — y − 8 = 12(x − 2).f'(x) = 3x². f'(2) = 12. Tangent: y − 8 = 12(x − 2) → y = 12x − 16.
f(x) = x³ − 3x has critical points at: — x = ±1.f'(x) = 3x² − 3 = 3(x²−1) = 0 → x = ±1.
For f(x) = x³ − 3x: classify x = 1: — Local min.f''(x) = 6x. f''(1) = 6 > 0 → local minimum at x = 1.
For f(x) = x³ − 3x: classify x = −1: — Local max.f''(−1) = −6 < 0 → local maximum at x = −1.
Absolute (global) maximum on a closed interval [a,b] must be found by checking: — Critical points AND endpoints.The Extreme Value Theorem guarantees a max and min on [a,b]. Check all critical points in (a,b) AND the endpoints a and b.
f''(c) = 0 implies: — The Second Derivative Test is inconclusive at c.If f''(c) = 0, the Second Derivative Test is inconclusive — c could be a min, max, or inflection point.
The first derivative test: if f'(x) changes from positive to negative at c, then c is: — Local maximum.Positive f' then negative f' means the function increased then decreased — so c is a local maximum.
Linear approximation: f(x) ≈ f(a) + f'(a)(x − a) near x = a. This is called: — Linearization of f at a.The linear approximation (linearization) uses the tangent line to approximate f near a point.
A critical number c is a global minimum on [a,b] if f(c) ≤ f(x) for: — All x in [a,b].A global minimum on [a,b] satisfies f(c) ≤ f(x) for ALL x in the closed interval [a,b].
The normal line to a curve at a point is: — Perpendicular to the tangent line.The normal line is perpendicular to the tangent line at the point of tangency.
If f'(a) = m, the normal line at (a, f(a)) has slope: — −1/m.Perpendicular lines have slopes that are negative reciprocals: m_normal = −1/m.
A ball thrown upward has position s(t) = −16t² + 64t. Maximum height occurs when: — t = 2 s.v(t) = s'(t) = −32t + 64 = 0 → t = 2 s. Both options A and D are equivalent.
At t = 2 s, the ball's maximum height is: — 64 ft.s(2) = −16(4) + 64(2) = −64 + 128 = 64 ft.