When the math has no clean answer, approximate it. These methods are short, mechanical and show up as quick points on the exam.
Choose any section below to get a free practice problem. Work it with the FE Reference Handbook open, tap your answer, then check the solution.
Bisection halves an interval that brackets a root until the interval is small enough.
Bisection on f(x) = x2 − 3 starts with [1, 2]. What is the second midpoint?
First midpoint 1.5: f(1.5) = −0.75 < 0 and f(2) = 1 > 0, so the root is in [1.5, 2]. Second midpoint: 1.75.
Common trap: (D) keeps the wrong half. Keep the half where the sign changes.
Numerical MethodsNewton’s methodbisectionThat was 1 of 3 problems in this section. The full course has the lesson, a worked example and every problem. Try all of it free for 3 days →
When you can’t integrate a function exactly, or you only have data points, approximate the area.
Estimate the integral of x2 from 0 to 2 with Simpson’s rule and h = 0.5.
(0.5/3)[0 + 4(0.25) + 2(1) + 4(2.25) + 4] = (0.1667)(16) = 2.667, exact for a parabola.
Common trap: (A) is the trapezoidal answer.
trapezoidal ruleSimpson’s ruleThat was 1 of 3 problems in this section. The full course has the lesson, a worked example and every problem. Try all of it free for 3 days →
Euler’s method steps along the slope: yn+1 = yn + h f(xn, yn).
One Euler step with h = 0.1 for y′ = −2y, y(0) = 1, gives y(0.1) ≈
y1 = 1 + 0.1(−2)(1) = 0.800.
Common trap: (C) is the exact value e−0.2.
Euler’s methodnumerical solutionThat was 1 of 3 problems in this section. The full course has the lesson, a worked example and every problem. Try all of it free for 3 days →
Finite differences estimate derivatives from nearby values.
Estimate the derivative of f(x) = x3 at x = 2 with a forward difference and h = 0.1.
(f(2.1) − f(2))/0.1 = (9.261 − 8)/0.1 = 12.61. Less accurate than a central difference, which gives 12.01 (exact: 12).
finite differenceforward differencecentral differenceThat was 1 of 3 problems in this section. The full course has the lesson, a worked example and every problem. Try all of it free for 3 days →
Algorithm questions hand you a few lines of pseudocode or a flowchart and ask what comes out, so build a trace table: one column per variable, one row per pass, updated in the exact order the statements are written.
What does this pseudocode print? (I MOD 3 is the remainder when I is divided by 3.) C = 0; FOR I = 1 TO 20: IF I MOD 3 = 0 THEN C = C + 1; END IF; END FOR; PRINT C
The IF is true for I = 3, 6, 9, 12, 15 and 18, so C goes up six times: 6.
Common trap: (C) adds I instead of 1, summing the multiples of 3 rather than counting them.
Flow ChartPseudocodeAlgorithmsThat was 1 of 4 problems in this section. The full course has the lesson, a worked example and every problem. Try all of it free for 3 days →
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