Continuity of a Function

Section 2.5: Continuity of a Function Removable and non-removable Discontinuities At At✗ 1.34 : =2 f- (2) UñtlÑ=1 : f- (1) = LHL=0 ? I tenth = ? 12111=4 µñtH'DN ✗→ I ✗→ I ' f is discontinuous at . . ✗→ 2 . = . ) his f- 1×7=15-12 ✗-12 + is discontinuous et −𝑥 + 1, 1, 2, 𝑓 (𝑥 ) = 𝑥 − 1, −𝑥 + 5, { 0.5, 0≤𝑥<1 1≤𝑥<2 𝑥=2 2<𝑥≤3 3<𝑥<4 𝑥=4 f is continuous at Example 2/ Page 66, 14th Edition, Thomas: Properties of Continuous Functions If k is a real number and 𝑓 and 𝑔 are continuous at 𝑥 = 𝑐, then the following functions are also continuous at 𝑐. 1. Scalar multiple: 𝑘𝑓 2. Sum and difference: 𝑓 ± 𝑔 r 3. Product: 𝑓𝑔 𝑓 4. Quotient: 𝑔 , 𝑖𝑓 𝑔(𝑐) ≠ 0 Continuity of a Composite function: If a function 𝒇 is continuous at 𝑐 and 𝑔 is continuous at 𝑓(𝑐 ), then the composite function given by (𝑔° 𝑓)(𝑥 ) = 𝑔(𝑓(𝑥 )) is continuous at c. Note: The following types of functions are continuous at every point in their domains. 1. Polynomial functions: 𝑝(𝑥 ) = 𝑎𝑛 𝑥 𝑛 + 𝑎𝑛−1 𝑥 𝑛−1 + ⋯ + 𝑎1 𝑥 + 𝑎0 2. Rational functions: 𝑓 (𝑥 ) = 𝑞(𝑥) , 𝑞(𝑥) ≠ 0 3. Radical functions: 𝑔(𝑥 ) = √𝑥 𝑝(𝑥) 𝑛 4. Exponential functions, Logarithmic functions, Absolute value functions Removable and Non-removable Discontinuities: A discontinuity at 𝑥 = 𝑐 is said to be removable if 𝑓 can be made continuous at 𝑥 = 𝑐 by redefining 𝑓(𝑐 ). f- ④ is Md¥ñed y= J= " DHE knife ) xsc it ' Examples: ya leinflx) { ✗ ±! .2_, f- flu xse .zy=Y¥'* ✗ +3.1¥ ✗ +3 , ✗ 71 Inuk /¥I,fH ' -12,5-1 Removab Mmremovab Find the x-values (if any) at which 𝑓 is not continuous. Which of the discontinuities are removable? - A Removable 1. 𝑓 (𝑥 ) = Uni 𝑥 2 −3𝑥−10 𝑥+2 discontinuity 𝑥 ≠ −2 - flx) f is at discontinuous ✗ g- (→ = -2 Not defined is . =uñY÷):ca÷ ✗→ -2 ✗→ = -2 Ani ✗→ → hai = ✗ → -2 (-1-5) =①_y ✗ 1=1×1 , 1=-2 = -77 2𝑥 − 4, 𝑥 ≤ 3 2. ℎ(𝑥 ) = { 2 𝑥 − 2𝑥, 𝑥 > 3 ✗ = -2 KIX = hL3) his = ✗→ −2𝑥 + 3, 𝑥 2, , leni (2×-4) = 3hl = zt f- ( x) f- ( 17 = him fix ✗ →I 1 = I } 31-42-2×7=3 leni non-removable A continuous at 2 . ✗→ 𝑥<1 𝑥≥1 = 3- ✗→ discontinuity • is 73 ✗ =3hnihl✗DM his h 1×7 ✗ → £3 ✗→ 3 2 = = RHL 3. 𝑓 (𝑥 ) = { ✗ ✗ , F- 2X at k¥ LHL { 2×-4 ✗ =/ 𝑥 + 1, 4. 𝑓(𝑥 ) = {𝑥 2 − 3𝑥 + 4 5 − 𝑥, 𝑖𝑓 𝑥 < 1 𝑖𝑓 1 ≤ 𝑥 ≤ 3 𝑖𝑓 𝑥 > 3 Examples: Use the intermediate Value Theorem to show that A) The polynomial function, 𝑓 (𝑥 ) = 𝑥 3 + 2𝑥 − 1 has a zero in the interval [0, 1]. f- (O ) f- (1) By = IVT -1 , and lies 0 -1 = 2 must f2 . C -1 between there must between every assume exist 0 and at and I 2 value between . least one ✗ - value such that f( c) =0 . B) Verify that the IVT applies to the indicated interval and find the value of c guaranteed by the theorem. 𝑓(𝑥 ) = 𝑥 3 − 𝑥 2 + 𝑥 − 2, [0, 3], 𝑓(𝑐 ) = 4 f- continuous function is a internal f-C) = f- (3) Since am ✗ - 19 closed . o between f- ( ) lies value f- 47 a -2 = 4 [ 0, 3) ones = 4 c . between and f-(3) 0 and 3 , there exists smh that