Simplify a ln() equation, determine a one-to-one function, and find a point that lies on the graph.?

Question

1) Simplify 2e^3ln(x+1)

2)Determine whether or not the following function is one-to-one. If it is one-to-one, write an equation for the inverse in the form y = f^-1 (x).
f(x) = (x+5)/(2x-3)

3) If the point (5,13) lies on the graph of the equation y =2k(x-3)^2 +5, then k equals...?

Could someone please explain how do to each of these? Thank you.

Answer 1

1) first, notice that e^ln B = B. B can be a number or a more complicated expression. So we have
2e^(3ln(x+1)) = 2(x+1)^3 = 2(x^3 + 2x^2 + 2x + 1) =
2x^3 + 4x^2 + 4x + 2

2. To determine if the function is 1:1, set f(x1) = f(x2), simplify, and see if it follows that x1 must equal x2. SInce this is a first degree rational function, it probalby would be 1:1. If there were squares or even powers in there you'd have instances whrere a postitive x1 and a negative x2 would give the same function value. Another way to tell if it's 1:1 is to graph it (easy enough to do on modern calculators) and see if a horozontal line goes thru it more than once. Looking at the graph, it appears that this function is 1:1.
To find the inverse, set y = f(x), interchange the variables, and solve for y. Here goes ...
y = (x + 5)/(2x - 3)
interchange x and y
x = (y + 5)/(2y - 3)
solve for y
x(2y - 3) = (y + 5)
2xy - 3x = y + 5
2xy - y = 3x + 5
y(2x - 1) = (3x + 5)
y = (3x + 5)/(2x - 1) -- there's the inverse


3) Substitute 13 for y and 5 for x 13 = 2k(5-3)^2 + 5
subtract 5 and get
8 = 2k (2)^2 divide by 2 and get
8 = 8k so k = 1

Answer 2

1) Keep in mind that the functions
f(x) = e^x and f(x) = ln(x) are inverses of each other. That means they cancel each other out to the functional identity "x". e^(lnx) = x, and ln(e^x) = x.

If we take your example,

2e^[3 ln(x+1)]

Your first step would be to apply the following log property:
log[base b](a^c) = c*log[base b](a)

This tells us that whenever we have an exponent inside of a log, we can take the exponent outside of the log as a non-exponent. Similarly, in the reverse direction, we can take c and put it BACK inside the log as a power. That's what we're going to do.

2e^(ln [(x+1)^3]

Now that we're taking e to the power of ln, and we know they are functional inverses of each other, we can cancel them out, leaving us with

2(x+1)^3

2. A function is one-to-one if the inverse exists. From a visual standpoint (i.e. looking at the graph), how to tell that a function is one-to-one is if it passes the horizontal line test (i.e. any horizontal line in any part of the graph will go through at most one point).

Algebraically, you have to set y = f(x) and then swap the terms.

f(x) = (x + 5) (2x - 3)

At this point, we can already tell it is a parabola, which are NEVER one-to-one. But we can still algebraically solve for y

y = (x + 5) (2x - 3). Now, swap x and y, and solve for y.
x = (y + 5) (2y - 3). Expand it out, to give you
x = 2y^2 - 3y + 10y - 15
x = 2y^2 + 7y - 15. We can move the x over, to get
0 = 2y^2 + 7y - 15 - x. Just to combined the last two terms,
0 = 2y^2 + 7y - (15+x). Now we can use the quadratic formula.

y = [-7 +/- sqrt(49 - 4(2)(15 + x))]/4
{This is already indicative of two solutions, but we'll simplify anyway}

y = [-7 +/- sqrt(49 - 120 - 8x)]/4
y = [-7 +/- sqrt(-71 - 8x)]/4

When we find an inverse of a function, we want ONE solution. Therefore, this is not one-to-one.

3. y = 2k(x - 3)^2 + 5

If (5,13) lies on the graph, all you have to do is plug in x = 5 and y = 13. This gets you

13 = 2k(5 - 3)^2 + 5

And you just solve for k.

13 = 2k(2)^2 + 5
13 = 8k + 5
8 = 8k
k = 1.

Answer 3

Your point is (2, √3) g(x) = √(x² - 1) (assuming this is all under the radical) Set g(x) = √3 from your point, and plug in 2 where there's an x. See if the equation is true. √3 = √(2² - 1) √3 = √(4 - 1) √3 = √3 Since the equation is true, the point lies on the graph. If your point had been (2, √5), it wouldn't have worked out and would not be on the graph.

Answer 4

One To One Function Equation

Answer 5

1) 2(x+1)^3

2)one-to-one


3)substitute (5,13) for x & y in the equation

8k+5=13

k=1

Answer 6

7-8