Category: 0. BLOG

For f(x)=x4−4x2f(x)=x4−4×2, which xx-values are inflexion candidates from f′′(x)=0f′′(x)=0? A x=±13x=±31 B x=±1x=±1 C x=±2x=±2 D x=±23x=±32 Explanation f′′(x)=12×2−8=4(3×2−2)f′′(x)=12×2−8=4(3×2−2). Setting

For f(x)=x3−3xf(x)=x3−3x, the point of inflexion occurs at A x=0x=0 B x=1x=1 C x=−1x=−1 D x=3x=3 Explanation f′′(x)=6xf′′(x)=6x. Setting f′′(x)=0f′′(x)=0

For a twice-differentiable function, the condition “f′(x)f′(x) increasing on an interval” directly implies A f′′(x)≤0f′′(x)≤0 B f′(x)=0f′(x)=0 C f′′(x)≥0f′′(x)≥0 D

While sketching y=f(x)y=f(x), the interval where f′′(x)>0f′′(x)>0 is mainly used to show A Downward bending B No curve change C

When a function is concave up on an interval, what is true about f′′(x)f′′(x) there? A f′′(x)0 D f′(x)=0f′(x)=0 Explanation

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