Author: Study Clue

If f′(x)>0f′(x)>0 for x≠0x=0 and f′(0)=0f′(0)=0, then f(x)f(x) is A Strictly decreasing B Constant function C Strictly increasing D Periodic

For f(x)=x2−4x+1f(x)=x2−4x+1, the minimum value is A B. −4−4 B C. 11 C A. −3−3 D D. 44 Explanation f(x)=x2−4x+1f(x)=x2−4x+1

For f(x)=x3−3xf(x)=x3−3x, the critical points are at A x=±3x=±3 B x=0,3x=0,3 C x=±1x=±1 D x=±3x=±3 Explanation Compute f′(x)=3×2−3=3(x2−1)f′(x)=3×2−3=3(x2−1). Set f′(x)=0f′(x)=0

In monotonicity testing, the interval points are usually split using A Discontinuity points only B Zeros of ff C Roots

If f′(x)>0f′(x)>0 for every xx in an interval, what can you conclude about f(x)f(x) on that interval A A. Strictly

If lim⁡x→0sin⁡(ax)sin⁡(2x)=3limx→0sin(2x)sin(ax)=3, then a equals A a = 6 B a = 3 C a = 2 D a =

Evaluate lim⁡x→0sin⁡2xsin⁡5xlimx→0sin5xsin2x A 5/2 B 10 C 0 D 2/5 Explanation For small x, sin⁡kx∼kxsinkx∼kx. So sin⁡2xsin⁡5x∼2x5x=2/5sin5xsin2x∼5x2x=2/5. This uses standard

Evaluate lim⁡x→2×2−4x−2limx→2x−2×2−4 A 0 B 4 C 2 D 8 Explanation Factor x2−4=(x−2)(x+2)x2−4=(x−2)(x+2). Cancel (x−2) for x≠2, so limit becomes

Which statement defines lim⁡x→af(x)=Llimx→af(x)=L A f(x) near L B f(a) equals L C x equals a D f(x) constant Explanation

If LHL = RHL = 5 at x=a, then limit at a equals A 5 B Does not exist C