a million) remember that the applications f(x) = e^x and f(x) = ln(x) are inverses of another. meaning they cancel one yet another out to the clever identity "x". e^(lnx) = x, and ln(e^x) = x. If we take your occasion, 2e^[3 ln(x+a million)] Your first step could be to stay with right here log assets: log[base b](a^c) = c*log[base b](a) This tells us that on each occasion we've an exponent interior a log, we can take the exponent outdoors of the log as a non-exponent. further, interior the opposite course, we can take c and positioned it decrease back interior the log as a ability. that's what we will do. 2e^(ln [(x+a million)^3] Now that we are taking e to the flexibility of ln, and all of us understand they are clever inverses of another, we can cancel them out, leaving us with 2(x+a million)^3 2. A function is one-to-one if the inverse exists. From a seen perspective (i.e. finding on the graph), the thank you to tell that a function is one-to-one is that if it passes the horizontal line attempt (i.e. any horizontal line in any area of the graph will flow by using at maximum one element). Algebraically, you're able to desire to set y = f(x) and then exchange the words. f(x) = (x + 5) (2x - 3) At this element, we can already tell that's a parabola, that are by no ability one-to-one. yet we can nevertheless algebraically remedy for y y = (x + 5) (2x - 3). Now, exchange x and y, and remedy for y. x = (y + 5) (2y - 3). develop it out, to grant you x = 2y^2 - 3y + 10y - 15 x = 2y^2 + 7y - 15. we can flow the x over, to get 0 = 2y^2 + 7y - 15 - x. just to mixed the final 2 words, 0 = 2y^2 + 7y - (15+x). Now we can use the quadratic formula. y = [-7 +/- sqrt(40 9 - 4(2)(15 + x))]/4 {it fairly is already indicative of two innovations, yet we will simplify besides} y = [-7 +/- sqrt(40 9 - a hundred and twenty - 8x)]/4 y = [-7 +/- sqrt(-seventy one - 8x)]/4 while we come across an inverse of a function, we want ONE answer. as a result, it fairly is not one-to-one. 3. y = 2k(x - 3)^2 + 5 If (5,13) lies on the graph, all you're able to desire to do is plug in x = 5 and y = 13. This gets you 13 = 2k(5 - 3)^2 + 5 and you in basic terms remedy for ok. 13 = 2k(2)^2 + 5 13 = 8k + 5 8 = 8k ok = a million.
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a million) remember that the applications f(x) = e^x and f(x) = ln(x) are inverses of another. meaning they cancel one yet another out to the clever identity "x". e^(lnx) = x, and ln(e^x) = x. If we take your occasion, 2e^[3 ln(x+a million)] Your first step could be to stay with right here log assets: log[base b](a^c) = c*log[base b](a) This tells us that on each occasion we've an exponent interior a log, we can take the exponent outdoors of the log as a non-exponent. further, interior the opposite course, we can take c and positioned it decrease back interior the log as a ability. that's what we will do. 2e^(ln [(x+a million)^3] Now that we are taking e to the flexibility of ln, and all of us understand they are clever inverses of another, we can cancel them out, leaving us with 2(x+a million)^3 2. A function is one-to-one if the inverse exists. From a seen perspective (i.e. finding on the graph), the thank you to tell that a function is one-to-one is that if it passes the horizontal line attempt (i.e. any horizontal line in any area of the graph will flow by using at maximum one element). Algebraically, you're able to desire to set y = f(x) and then exchange the words. f(x) = (x + 5) (2x - 3) At this element, we can already tell that's a parabola, that are by no ability one-to-one. yet we can nevertheless algebraically remedy for y y = (x + 5) (2x - 3). Now, exchange x and y, and remedy for y. x = (y + 5) (2y - 3). develop it out, to grant you x = 2y^2 - 3y + 10y - 15 x = 2y^2 + 7y - 15. we can flow the x over, to get 0 = 2y^2 + 7y - 15 - x. just to mixed the final 2 words, 0 = 2y^2 + 7y - (15+x). Now we can use the quadratic formula. y = [-7 +/- sqrt(40 9 - 4(2)(15 + x))]/4 {it fairly is already indicative of two innovations, yet we will simplify besides} y = [-7 +/- sqrt(40 9 - a hundred and twenty - 8x)]/4 y = [-7 +/- sqrt(-seventy one - 8x)]/4 while we come across an inverse of a function, we want ONE answer. as a result, it fairly is not one-to-one. 3. y = 2k(x - 3)^2 + 5 If (5,13) lies on the graph, all you're able to desire to do is plug in x = 5 and y = 13. This gets you 13 = 2k(5 - 3)^2 + 5 and you in basic terms remedy for ok. 13 = 2k(2)^2 + 5 13 = 8k + 5 8 = 8k ok = a million.
By trial and error when x = 3, y = √1 + 5 = 1+ 5 = 6
C. (3,6)