Solve the problem.
-Estimate graphically the local maximum and local minimum of $f ( x ) = 0.0001 x ^ { 4 } - 0.1 x ^ { 2 } + 0.1$.
A) Local maximum: $0.1$; local minima: $\pm 24.9$
B) Local maximum: 0 ; no local minimum
C) Local maximum: 0.1; no local minimum
D) Local maximum: 0; local minima: $\pm 22.4$

Question

Quizplus · Accepted Answer

The function $f(x) = 0.0001x^4 - 0.1x^2 + 0.1$ is a quartic polynomial. Its graph is symmetric about the y-axis because it only contains even powers of x. The term $-0.1x^2$ suggests the graph will have a local maximum at $x = 0$, since the coefficient of $x^2$ is negative, indicating a downward opening parabola for the quadratic part of the function. The $0.0001x^4$ term ensures that as $x$ moves away from 0, the function will eventually turn upwards, indicating the presence of two local minima on either side of the y-axis. The constant term $0.1$ shifts the entire graph upwards by 0.1 units. Without exact calculation, the exact positions of the local minima cannot be determined precisely, but option D correctly identifies the presence of a local maximum at $x = 0$ and local minima on both sides of the y-axis, which aligns with the expected behavior of the function's graph.

Solve the Problem f(x)=0.0001x4−0.1x2+0.1f ( x ) = 0.0001 x ^ { 4 } - 0.1 x ^ { 2 } + 0.1f(x)=0.0001x4−0.1x2+0.1

Solve the Problem $f ( x ) = 0.0001 x ^ { 4 } - 0.1 x ^ { 2 } + 0.1$