This document contains solutions to the questions in MAT4MDS Assignment 2, 2019. It includes solving equations for x, proof of logarithmic equations, graphing equations, and finding the inverse function of F(x).
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Using trial and error method 𝒙= 𝟏. 𝟏𝟕𝟎𝟖𝟑 b)Proof i)loga(b)logb(a)= 1 logb(a)=1 logab 1 =1 logab×1 logba×(loga(b)logb(a)) 1 = (1 logab× loga(b)) × (1 logba× logb(a)) 1 = 1 × 1 = 1 ii)log1 b (x)= − logb(x) logb−1(x)= − logb(x)use of change of base rule ln(x) ln(b−1)= −ln(x) ln(b) ln(x) −ln(b)= −ln(x) ln(b) ln(x) ln(b)= −ln(x) ln(b) c)Graphing the following equations on the same axis, c=8 𝑦= 𝑥,(1 5) 𝑥 , (1 8) 𝑥
d)Estimating the slope of the red line The graph is logarithmic on y-axis s =∆y ∆x=∆W ∆nat(1,1)and (2,10) Where W total number of websites, and n the number of year s=∆W ∆n=1 2−1=1∆W = 1for one log cycle ∴s = 1 yintercept= −10 Thus, y = −10(10)x The relationship between W and n is Exponential relationship Question 2 F:[0, 1}→ ℝF(x)= 1 −(1 − xa)b
a)The range for F is [0, 1] b)F(x)= 1 −(1 − xa)b f = 1 − xag = xb c)a = 5,b == 1 8 𝑔°𝑓(𝑥)=(1 − 𝑥𝑎)1 8 Reflecting on x-axis Shifting vertically up by +1
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d)Both a and b are scale parameters since they both change the size and the shape of the function. e)F−1(x) y = 1 −(1 − xa)b (1 − xa)b= 1 − y ln(1 − xa)=ln(1 − y) b 1 − xa= eln(1−y) b⟹xa= 1 − e ln(1−y) b⟹aln(x)= ln (1 − e ln(1−y) b) x = e ln(1−e ln(1−y) b) a ∴F−1(x)= e ln(1−e ln(1−x) b) a Range:[0, 1] Domain: [0, 1]