Download Mathematics, Calculus Year 1 Inverse functions Please solve each

Mathematics, Calculus
Year 1
Inverse functions
Please solve each problem with a detailed solution showing each step to solve the problem. Since
the symbols confuse me at times please use “baby” math to show how to get from the start to the
end. I understand the book in some ways, but the more I see completed the better I can think
about the rest of the problems I need to do. Also, if you type it all back to me, not a scanned print
of your help, please keep things separated and neat please. Also, I am a returning adult student you may know this stuff inside and out, but I do not. Thanks!
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Inverse function problems:
For the following graph the given functions on a computer screen, how are these graphs related?
1) Y=2^x, y=e^x, y=5^x, y=20^x
The graph of each of the function is draw on the following graph.
By looking at the graph, we see that each all the functions are exponential functions.
2) Y=3^x, y=10^x, y=(1/3)^x, y=(1/10)^x
From the graph we see that 3^x and (1/3)^x are inverse to each other. Similarly 10^x and
(1/10)^x are inverse to each other.
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Make a sketch of the function.
7) y=4^x-3
Solution:
First draw the graph of y = 4^x.
If you subtract 3 from the function, then the graph 4^x -3 will be shifted 3 units down.
8) y=-2x^-x
Solution:
Here the basic function is y=x^(-x).
The graph of x^(-x) is the following.
Then multiply the function by 2, the graph is stretches by 2 units.
Now, multiply 2x^(-x) by -1. Then the graph is reflected about x axis.
9) y=3-e^x
Solution:
Here the base function is e^x.
The graph of e^x is,
Then multiply e^x by -1. Then the graph is reflected about x axis.
The next step is, you are going to add 3. –e^x + 3 is nothing but 3-e^x. So you are actually adding
3.
So the graph is shifted 3 units up.
13) Starting with the graph of y=e^x, write the equation of the graph that results from.
a) shifting two units downwards.
To shift the graph two units down, you have to subtract 2 from the function.
So the function is y = e^x -2.
b) shifting two units to the right.
To shift the graph two units up, you have to add 2 from the function.
So the function is y = e^x + 2.
c) reflecting about the x axis.
Reflections in the coordinate axes of the graph of y = f(x) are represented as follows.
1. Reflection in the x axis : y = -f(x).
2. Reflection in the y axis : y = f(-x).
So to get reflected about x axis, change the sign of the whole function.
So y =
 e x is reflection of e x about x axis.
e) ) reflecting about the x axis and then about the y axis.
To get reflected about y axis, change the sign of x in the function.
That is y =
e  x is reflection of e x about y axis
Find the domain of the functions:
Domain: The set of x values for which the function is defined.
To find the domain, we find out the values for which the function is not defined. All the values
except for which the function is not defined is the domain of the function.
For a rational function: The denominator of a rational function cannot be zero.
So we solve for the zeros of the denominator. All values except zeros of the denominator is the
domain of a rational function.
a) f(x)=1/(1+e^x)
Solution:
To find the domain of a ration function, equate the denominator to zero.
So, 1+e^x = 0.
 e^x = -1.
Exponential function always takes positive value.
There exist no such number. So the denominator never becomes zero.
That means the domain is all real numbers.
b) f(x) = 1/(1-e^x)
Solution:
Equate the denominator to zero.
1 – e^x = 0.
Add e^x both sides.
That gives, 1 = e^x.
The solution for this equation is x =0. That is e^x takes the value 1, when x=0.
So the domain is all numbers except 0.
c) g(t)=sin(e^-1)
This is a function of t. But there is no t on the right side.
So the function is a constant function. The domain of a constant function is “ all numbers”.
d) g(t) = SQRT(1-2^t)
Square root of a function is defined only if the value inside the square root is positive.
So the function is defined for positive values of (1-2^t).
That is (1-2^t)  0.
For this to be greater than 0, 2^t should be less than 1.
2^t is less than 1 only for negative values of t.
So the domain of the function is all negative numbers.
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19) Suppose the graphs of f(x)= x^2 and g(x)=2^x are drawn on a coordinate grid where the unit
of measurement is 1 inch. Show that at a distance of 2 feet to the right of the origin, the height of
the graph of f is 48ft but the height of the graph of g is about 265mi. (Please check that your
function f(x) must be equal f(x) = 2x.)
The height of the graph is nothing but the value of the y coordinate.
Note that the unit of measurement is 1 inch. So at 2feet = 24 inches. So 24 inches to the right of
the origin, the value of x = 24.
So f(x) = 2x  f(24) = 2(24)= 48inches.
and g(x) = 2^x  f(24) = 2^24 = 16777216inches  265 miles.
20) Compare the functions f(x)=x^10 and g(x)=e^x by graphing both f and g in several viewing
rectangles. When does the graph of g finally surpass the graph of f?
Draw the graph of f(x) = x^10 and g(x) = e^x on the same grid.
So, the graph of g surpass the graph of f at x = 1.
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Find the limits for the following
1) lim as x approaches infinity(1.001)^x
Solution:
lim
x
(1.001) x = (1.001)  = 
2) lim as x approaches infinity(e)^-2x
lim
x
e  2 x  e  2(  )  e  0
3) lim as x approaches infinity(e^3x – e^-3x)/(e^3x + e^-3x)
e 3 x  e 3 x
x   e 3 x  e 3 x
lim
If we apply limit as it is we will get in to trouble as there is
denominator, that will become infinity as x approaches infinity.
e 3 x from both the numerator and denominator.
e 3 x (1  e 6 x )
3x
3 x
 3x
because, e  e
 e 3 x  e 3 x e 3 x e 3 x  e 3 x  e 3 x e 6 x
6 x
e (1  e )
So, first factor out
e 3 x  e 3 x
e 3 x  e 3 x
e 3 x in both numerator and
=
1  e 6 x
1  e 6 x
Now, applying limit as x approaches infinity,
lim 1  e 6 x 1  e 

x   1  e 6 x 1  e 
= 1 since
e   0 .
4) lim as x approaches (pi/2)^+ (e^tanx)
lim
x  /2
e tan( x ) = e 

since the value of tan(x) as x approaches

from the right, is - 
2
= 0.
5) lim as x approaches 2^+ (e^(3/(2-x)))
3
lim
[e 2  x ]
x  2
As x approaches 2 from the right, the values of x are greater than 2. so the sign of (2-x) is
negative. So as x approaches 2 from the right, the limit of
So,
lim
x2

[e
3
2 x
3
is negative infinity.
2x
] = e  = 0.
6) lim as x approaches 2^- (e^(3/(2-x)))
lim
[e
x  2
3
2 x
]
As x approaches 2 from the left, the values of x are less than 2. so the sign of (2-x) is
positive. So as x approaches 2 from the left, the limit of
So,
lim
x  2
[e
3
2 x
] = e = 
Differentiate the functions:
1) f(x) =x^2e^2
Solution:
y = f(x) = x2 e2
d 2
d 2
(e )  e 2
(x )
dx
dx
 ( x 2 )(0)  (e 2 ) (2 x)
f ' ( x)  ( x 2 )
 2e 2 x
2) y=e^ax^3
Solution:
y = f(x) = x3 ea
3
2x
is infinity.
d a
d 3
(e )  e a
(x )
dx
dx
 ( x 2 )( 0)  (e a ) (3 x 2 )
f ' ( x)  ( x 3 )
 3e a x 2
3) f(u) =e^(1/u)
Solution:
f (u )  e (1 / u )
f ' (u )  e (1 / u ) (
1
)
u2
apply chain rule
4) f(t)=e^(tsin2t)
Solution:
f (t )  e (t sin 2t )
f ' (t )  e (t sin 2t ) (sin 2t  2 cos 2t * t )
apply chain rule
5) y=SQRT(1+2e^3x)
Solution:
y  1  2e 3 x
y' 
y' 
1
2 1  2e 3 x
3e 3 x
* 6e 3 x
1  2e 3 x
6) y=e^e^x
Solution:
y  ee
x
y '  ee e x
x
7) y= (ae^x+b)/(ce^x+d)
Solution:
y
(ae x  b)
(ce x  d )
dy (ce x  d )( ae x )  (ae x  b)(ce x )

dx
(ce x  d ) 2
8) y=e^x/(1+x)
Solution:
apply quotient rule
ex
1 x
(1  x) (e x )  (e x )
y' 
(1  x) 2
y

xe x
(1  x) 2
9) g(x)=SQRT(x)e^x
Solution:
g ( x) 
xe x
g ' ( x) 
1
ex  ex x
2 x
1  2x
 ex
2 x
10) y=e^(ktanSQRT(x))
Solution:
y  e ( k tan
x)
y '  e ( k tan
x)
(k sec x )
1
2 x
apply function of function rule.
11) y=SQRT(1+xe^-2x)
Solution:
y  1  xe2 x
y' 

1
2 x
2 1  xe
e  2 x (1  2 x)
* (e  2 x  2e  2 x x)
2 1  xe 2 x
Find the equations of the tangent line to the curve at the given point.
1) y=e^(2x) cos(pi)(x), (0,1)
Solution:
y  e 2 x cos x , (0,1)
dy
 (e 2 x )(  sin x  )  (cos x) (2e 2 x )
dx
 dy 
 (e 0 ) ( * 0)  1 * 2e 0
 
dx
  ( 0,1)
2
therefore m  2
Equation of the tan gent line is
y  y1  m ( x  x1 )
y  1  2 ( x  0)
y  2x  1
2) Find y’ if e^(x^2 times y)
Information is not correct.