a million) do no longer forget that the applications f(x) = e^x and f(x) = ln(x) are inverses of one yet another. meaning they cancel one yet another out to the smart identity "x". e^(lnx) = x, and ln(e^x) = x. If we take your occasion, 2e^[3 ln(x+a million)] Your first step may well be to stay with 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 exterior of the log as a non-exponent. in addition, interior the opposite path, we can take c and placed it returned interior the log as a potential. that's what we are going to do. 2e^(ln [(x+a million)^3] Now that we are taking e to the potential of ln, and all of us understand they're smart inverses of one yet 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), a 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 pass by using at maximum one ingredient). Algebraically, you may desire to set y = f(x) and then substitute the words. f(x) = (x + 5) (2x - 3) At this ingredient, we can already tell this is a parabola, that are by no skill one-to-one. yet we can nevertheless algebraically remedy for y y = (x + 5) (2x - 3). Now, substitute x and y, and remedy for y. x = (y + 5) (2y - 3). enhance it out, to furnish you x = 2y^2 - 3y + 10y - 15 x = 2y^2 + 7y - 15. we can pass 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 {this is already indicative of two suggestions, yet we are going to simplify besides} y = [-7 +/- sqrt(40 9 - one hundred twenty - 8x)]/4 y = [-7 +/- sqrt(-seventy one - 8x)]/4 as quickly as we come across an inverse of a function, we choose ONE answer. for this reason, this is not any longer one-to-one. 3. y = 2k(x - 3)^2 + 5 If (5,13) lies on the graph, all you may 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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Verified answer
B. 0, 5
Everything to the left of X=2 results in an imaginary part (sqrt(neg)). Everything at x=2 and to the right are real, and on the graph.
a million) do no longer forget that the applications f(x) = e^x and f(x) = ln(x) are inverses of one yet another. meaning they cancel one yet another out to the smart identity "x". e^(lnx) = x, and ln(e^x) = x. If we take your occasion, 2e^[3 ln(x+a million)] Your first step may well be to stay with 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 exterior of the log as a non-exponent. in addition, interior the opposite path, we can take c and placed it returned interior the log as a potential. that's what we are going to do. 2e^(ln [(x+a million)^3] Now that we are taking e to the potential of ln, and all of us understand they're smart inverses of one yet 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), a 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 pass by using at maximum one ingredient). Algebraically, you may desire to set y = f(x) and then substitute the words. f(x) = (x + 5) (2x - 3) At this ingredient, we can already tell this is a parabola, that are by no skill one-to-one. yet we can nevertheless algebraically remedy for y y = (x + 5) (2x - 3). Now, substitute x and y, and remedy for y. x = (y + 5) (2y - 3). enhance it out, to furnish you x = 2y^2 - 3y + 10y - 15 x = 2y^2 + 7y - 15. we can pass 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 {this is already indicative of two suggestions, yet we are going to simplify besides} y = [-7 +/- sqrt(40 9 - one hundred twenty - 8x)]/4 y = [-7 +/- sqrt(-seventy one - 8x)]/4 as quickly as we come across an inverse of a function, we choose ONE answer. for this reason, this is not any longer one-to-one. 3. y = 2k(x - 3)^2 + 5 If (5,13) lies on the graph, all you may 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.