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course Mth 173
9/27 17:35
text 07PROBLEM 1-8
Given f(x) = x^2 and g(x) = x + 1
(a) f(g(1))
g(1) = 1 + 1 = 2
and f(g(1)) = f(2) = 2^2 = 4
thus f(g(1)) = 4
(b) g(f(1))
f(1) = 1^2 = 1
and g(f(1)) = g(1) = 1 + 1 = 2
thus g(f(1)) = 2
(c) f(g(x))
g(x) = x + 1
and f(g(x)) = g(x) ^ 2 = ( x + 1 ) ^ 2 = x^2 + 2x + 1
thus f(g(x)) = x^2 + 2x + 1
(d) g(f(x))
f(x) = x^2
and g(f(x)) = f(x) + 1 = x^2 + 1
thus g(f(x)) = x^2 + 1
(e) f(t)g(t)
f(t) = t^2
and g(t) = t + 1
thus f(t)g(t) = f(t) * g(t) = t^2 * ( t + 1 ) = t^3 + t^2
PROBLEM 1-10
Given f(x) = e ^ x and g(x) = x ^ 2
(a) f(g(1))
g(1) = 1^2 = 1
and f(g(1)) = f(1) = e ^ 1 = e
thus f(g(1)) = e
(b) g(f(1))
f(1) = e ^ 1 = e
and g(f(1)) = g(e) = e ^ 2
thus g(f(1)) = e ^ 2
(c) f(g(x))
g(x) = x ^ 2
and f(g(x)) = f( x ^ 2 ) = e ^ ( x ^ 2 ) = e ^ ( 2x)
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x^2 is not generally equal to 2x, so e^(x^2) is not equal to e^(2x).
You appear to be confusing e^(x^2) with (e^x)^2, which would be e^(2x).
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(d) g(f(x))
f(x) = e ^ x
and g(f(x)) = g( e ^ x ) = (( e ^ x ) ^ 2 ) = e ^ (2x)
(e) f(t)g(t)
f(t) = e ^ t
and g(t) = t ^ 2
thus f(t) * g(t) = ( e ^ t ) * ( t ^ 2 )
PROBLEM 1-13
Given f(x) = x ^ 2 + 1
(a) f(t + 1) = (t + 1 ) ^ 2 + 1 = t^2 + 2t + 1 + 1 = t^2 + 2t + 2
(b) f( t^2 + 1) = ( t^2 + 1 ) ^ 2 + 1 = t^4 + 2*( t^2) + 1 + 1 = t^4 + 2(t^2) + 2
(c) f(2) = 2 ^ 2 + 1 = 4 + 1 = 5
(d) 2f(t) = 2 * ( t ^ 2 + 1 ) = 2( t^2 ) + 2
(e) (f(t))^2 + 1 = ( t ^ 2 + 1 ) ^ 2 + 1 = t ^ 4 + 2( t^2 ) + 1 + 1 = t ^ 4 + 2(t^2) + 2
PROBLEM 1-14
Given m(z) = z ^ 2
m( z + 1 ) = ( z + 1 ) ^ 2 = z^2 + 2z + 1
m(z) = z^2
thus m( z + 1 ) - m(z) = z^2 + 2z + 1 - z^2 = 2z + 1
PROBLEM 1-15
Given m(z) = z^2
m( z + h ) = ( z + h )^2 = z^2 + 2zh + h^2
m(z) = z^2
thus m( z + h ) - m(z) = z^2 + 2zh + h^2 - z^2 = 2zh + h^2
PROBLEM 1-16
Given m(z) = z^2
m(z) = z^2
m( z - h ) = ( z - h )^2 = z^2 - 2zh + h^2
thus m(z) - m( z - h ) = z^2 - z^2 + 2zh - h^2 = 2zh - h^2
PROBLEM 1-20
f(x) is a function whose result is temperature(℉) when the column of mercury in a particular thermometer is x inches long. Thus for the function f(x) we have x as length of mercury in inches and y as temperature (℉). Thus f-1(x) is the inverse function of f(x) which means that domain of f(x) in the range of f-1(x) and range of f(x) is domain of f-1(x). Thus f-1(75) in practical terms means the length of mercury in inches when the temperature if 75℉.
PROBLEM 1-21
(a)
Given f(x) = x^2
The graph stretched by a factor of 2 thus the function becomes
f(x) = 2*(x^2)
a vertical upward shift of 1 results in
f(x) = 2*(x^2) + 1
(b)
Given f(x) = x^2
The graph given a vertical upward shift of 1 results in
f(x) = x^2 + 1
and the graph next stretched by 2 units thus the resulting graph becomes
f(x) = 2*( x^2 + 1 ) = 2*(x^2) + 2
(c)
The 2 graphs obtained are not equal. The order of the operations is definitely important.
For the first order the graph first stretches and then shifts the function by 1 unit.
For the second order the graph first shifts to 1 units and then stretched by 2 units so the shifting of 1 unit also got doubled by 2. Thus for the first order the shift was only of 1 unit , but for the second order since the shift was doubled too , thus the shift turned out to be of 2 units which is the only difference in both the graphs.
PROBLEM 1-22
(a) y = f(x) + 3
The given function can be obtained by shifting the given f(x) graph by 3 units upward.
The graph of f(x) + 3 will have the same slopes as that of the function f(x)
For the new graph the x intercept will become 0 and y intercept = y intercept for f(x) + 3 = 0 + 3 = 3
There is 1 asymptote to the graph f(x) which is y = 0, thus the asymptote for f(x) + 3 will become y = 0 + 3 = 3.
Thus the asymptote for f(x) + 3 is y = 3
(b) y = 2f(x)
The graph is stretched by a unit of 2. Each value of function f(x) is stretched by multiplying each value by a factor of 2. For example the point ( 2 , 2 ) becomes ( 2 , 4 ) and ( -3 , 1 ) becomes ( -3 , 2 ).
The y intercept and the asymptote remains unaffected.
(c) y = f( x + 4 )
We know that for a function f(x) , f( x - h ) shifts the graph by h units to the right.
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Good.
The simple way to understand this is that x - h must be h units greater to give the same result as x.
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Comparing the functions we get that h = -4 which means that the graph is shifted by -4 units to the right that means a units of 4 to the left. Example of points on the graph , the point ( 0 , 0 ) becomes ( -4 , 0 ) and the point ( 2 , 2 ) becomes ( -2 , 2 )
The asymptote remains unaffected
(d) y = 4 - f(x)
The first thing to do is to obtain the graph of - f(x) which is taking the negative of each value of f(x) and which can be obtained by reflecting the graph f(x) about the x axis. Some points example ( 2 , 2 ) becomes ( 2 , -2 ) and the point ( -3 , 1 ) becomes ( -3 , -1).
Next the function become 4 - f(x) , the graph for the same can be obtained by shifting the graph of - f(x) by 4 units vertically upwards. The values of the - f(x) are added each by 4 units. Thus points on - f(x) which are ( 2 , -2 ) and ( -3 , -1 ) become ( 2 , 2 ) and ( -3 , 3 ) respectively on the graph of 4 - f(x)
The asymptote to the function is y = 0. After the transformation the asymptote becomes y = 0 + 4 = 4
PROBLEM 1-23
For a function to be invertible
The function should be one to one that a vertical line parallel to x axis should not cut the graph at 2 or more points. The reason is if the line cuts the graph at 2 or more points, we have 2 values of x corresponding to 1 values of y. For an inverse function the domain of the f(x) function becomes the range of the inverse function and the range of the f(x) function becomes the domain of the inverse function that is the x and y values interchange. Thus for the inverse function we would have 2 values of y for one value of x which is not valid for function. Thus for a function f(x) to be invertible it should be an one to one function.
For the given graph which is a parabola, we see that any horizontal line parallel to x axis above a particular value of x (the x value of the vertex) cuts the parabola at 2 points. Thus the function is a many to one function (that is not a one to one function) and the thus the function is not invertible.
PROBLEM 1-24
For a function to be invertible
The function should be one to one that a vertical line parallel to x axis should not cut the graph at 2 or more points. The reason is if the line cuts the graph at 2 or more points, we have 2 values of x corresponding to 1 values of y. For an inverse function the domain of the f(x) function becomes the range of the inverse function and the range of the f(x) function becomes the domain of the inverse function that is the x and y values interchange. Thus for the inverse function we would have 2 values of y for one value of x which is not valid for function. Thus for a function f(x) to be invertible it should be an one to one function.
For the given graph, let a and b be the y values of the critical points of the graph. Now any horizontal line y = c , ( c between a and b ) would intersect the graph at 3 points, thus the graph becomes a many to one function and thus is no longer an invertible function.
!!!!!!!!!! can we define the inverse of the function with domain and range obtained by eliminating that part of the function f(x) between y = a and y = b, by doing that the function becomes a one to one function.
PROBLEM 1-30
Given f(x) = x^4 - x^2 + 3
And f(-x) = (-x)^4 - (-x)^2 + 3 = x^4 - x^2 + 3
Since f(x) = f(-x) thus the function is even
PROBLEM 1-34
Given f(x) = x*( x^2 - 1 )
f(-x) = (-x) * ( (-x)^2 - 1 ) = (-x) * ( x^2 - 1 )
Since f(-x) = -f(x) thus the function is odd
PROBLEM 1-35
Given f(x) = e^x - x
f(-x) = e^(-x) - (-x) = 1 / e^x + x
Neither f(x) is equal to f(x) nor is f(x) equal to -f(-x)
Thus the function is neither ( neither odd nor even )
PROBLEM 1-37
Given h(x) = x^3 + 1
We know h(x) = f(g(x))
For x^3 + 1 , the first step is cubing, so the inside function is g(x) = x^3
The second step is adding 1 , so the outside function is f(x) = x + 1
Thus f(g(x)) = g(x) + 1 = x^3 + 1
PROBLEM 1-38
Given h(x) = sqrt( x^2 + 4 )
We know h(x) = f(g(x))
For sqrt(x^2 + 4) , the first step is squaring and adding 4 , so the inside function is g(x) = x^2 + 4
The second step is taking the square root , so the outside function is f(x) = sqrt(x)
Thus f(g(x)) = sqrt( g(x)) = sqrt ( x^2 + 4 )
PROBLEM 1-43
The values of f-1(x) are as follow
x = -7 and f-1(-7) = 2
x = 1 and f-1(1) = 7
x = 2 and f-1(2) = 6
x = 3 and f-1(3) = 1
x = 4 and f-1(4) = 4
x = 19 and f-1(19) = 3
x = 178 and f-1(178) = 5
the domain for f(x) is all real numbers from 1 to 7 , which is now the range for f-1(x) and the range for f(x) becomes the domain for f-1(x) only if the function is bijective.
Thus the domain of f-1(x) is { -7 , 1 , 2 , 3 , 4 , 19 , 178 }
PROBLEM 1-45
f(t) is the number of customers in Macy’s department store at t minutes past noon on December 18,2008
now let for example be 5 customers in the store after 3 mins. It is very likely that there will be the same number of customers after 10 mins or more. this makes the function a many to one function which means that for 2 values of x we have the same value of y. if a graph for the same is drawn we will see that there is a horizontal line parallel to the x axis cuts the graph at 2 or more points.
for the inverse function the range becomes the domain and the domain becomes the range ( that is x and y interchange values) , thus for one value of x we will have 2 values of y , which violates the rule of a function. basically when plotted, if a line parallel to y axis cuts the graph at 2 or more points the expression is not a function. since the inverse expression is not a function, f(t) is not invertible.
thus the function f(t) is not invertible.
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Very good.
Check my one note on the fact that e^(x^2) is not the same as e^(2x).
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