Curriculum · Course · Week 3 · فارسی
An inverse recovers an input from an output. It exists on a domain where the original function is one-to-one. A horizontal line must meet the graph at most once. Swapping the coordinates of each point reflects the graph across y = x. An exponential changes by a fixed multiplicative factor over equal input intervals. A logarithm asks which exponent produces a specified positive value. These tools let you undo a sensor calibration or solve for time in a growth or decay model.
Write y = f(x), solve for x, and exchange the variable names. Check both the inverse domain and a composition. For logarithms, require a positive argument before manipulating the equation. The notation f⁻¹ means inverse, while 1/f means reciprocal. In ax, the variable is in the exponent. In xa, it is in the base.
A sensor model is v = 2x + 1 for 0 ≤ x ≤ 4. Undo the calibration.
x = (v − 1)/2, with 1 ≤ v ≤ 9. At v = 7, obtain x = 3 and verify 2(3)+1 = 7.
Can y = x² be inverted on all real inputs?
Inputs 2 and −2 both produce 4. Restrict to x ≥ 0 to obtain f⁻¹(y) = √y for y ≥ 0. A domain restriction resolves the ambiguity.
A model follows v(t) = 6e−t/2. When does it reach half its initial value?
3 = 6e−t/2 gives ln(1/2) = −t/2, so t = 2 ln 2 ≈ 1.386 s. The negative exponent represents decay.
The blue point (1,3) becomes (3,1) on the orange graph. The gray line is the mirror. Compare the domain [0,4] with the inverse domain [1,9]. Both axes use the same scale so the reflection is meaningful.
At t=0 the value is 6 V. At 2 s it is about 2.207 V, not zero. The curve approaches zero gradually. Mark the half-value level at 3 V and compare its crossing time with 2 ln 2.
The curve passes through (1,0), (2,1), (4,2) and (8,3). Each doubling of the input adds one to the output. There is no point on this graph at x=0.
For each problem, write the domain or units before calculating. After checking the answer, change one numerical value and solve the variant without looking back.
To undo a function uniquely, different allowed inputs must produce different outputs. A one-to-one function is invertible onto its range. Its inverse swaps the original domain and range. Writing an inverse formula without these sets can conceal a second branch or an impossible input. The identity f⁻¹(f(x))=x is asserted only for x in the original domain.
Suppose a calibrated sensor follows v=1+0.5x² for 0≤x≤4. Solve v−1=0.5x² to obtain x=√(2(v−1)). Choose the positive root because the original input is nonnegative. The output range is 1≤v≤9, which becomes the inverse domain. For v=5.5, the inferred input is 3. Substitution gives 1+0.5(3²)=5.5. If negative x were also allowed, the same reading would correspond to ±3 and the inverse would not be unique. The reciprocal 1/(1+0.5x²) is a different function altogether.
In y=A bˣ with b>0 and b≠1, increasing x by one multiplies the output by b. In y=A ekt, a time increment Δt multiplies the output by ekΔt. The coefficient k has units of inverse time, making the exponent dimensionless. The special base e≈2.71828 is the natural base used in continuous-time models. Its calculus properties come later; here it provides a convenient way to parameterise growth and decay.
For positive u and v, logarithms turn multiplication into addition: ln(uv)=ln u+ln v, ln(u/v)=ln u−ln v, and ln(uʳ)=r ln u. They do not split sums. Change of base follows by taking natural logs of bʸ=x, giving log_b(x)=ln x/ln b. Real logarithms require positive arguments and a positive base other than one.
A signal obeys v(t)=8 e−t/3 V, with t measured in seconds. Find when it reaches 1 V. Divide by 8, take the natural logarithm and solve: ln(1/8)=−t/3, so t=3 ln 8≈6.238 s. The time constant is 3 s, and at that time the output is 8/e≈2.943 V, not zero. The half-value time is 3 ln 2≈2.079 s. These are two different landmarks on the same curve. This is a mathematical decay model; component-level RC theory is not required yet.
Solve ln(x−1)+ln(x+1)=ln 8. Before combining, require x>1. Then ln(x²−1)=ln 8, hence x²=9. The algebraic candidates are ±3, but only x=3 satisfies the original log domains. Substituting x=−3 would ask for logs of negative numbers in this real-valued problem.
Complete problems 1–6. Problems 7–8 are optional extensions.
Common mistakes: ignoring the inverse branch, taking a log of a dimensional voltage instead of a voltage ratio, treating the time constant as the half-value time, and accepting every root of a transformed equation.
An original summary aligned with these textbook topics. References are optional further reading.