The functions f and g are defined by
f : x ↦ e^x + 2,
x ∈ ℝ
g : x ↦ ln x,
x > 0
(a) State the range of f - Edexcel - A-Level Maths Pure - Question 6 - 2012 - Paper 5
Question 6
The functions f and g are defined by
f : x ↦ e^x + 2,
x ∈ ℝ
g : x ↦ ln x,
x > 0
(a) State the range of f.
(b) Find fg(x), giving your answer in its simplest ... show full transcript
Worked Solution & Example Answer:The functions f and g are defined by
f : x ↦ e^x + 2,
x ∈ ℝ
g : x ↦ ln x,
x > 0
(a) State the range of f - Edexcel - A-Level Maths Pure - Question 6 - 2012 - Paper 5
Step 1
State the range of f.
96%
114 rated
Only available for registered users.
Sign up now to view full answer, or log in if you already have an account!
Answer
The function f(x) = e^x + 2 is defined for all real numbers x. Since e^x takes all positive values and adding 2 shifts this upward, the range of f is (2, ∞).
Step 2
Find fg(x), giving your answer in its simplest form.
99%
104 rated
Only available for registered users.
Sign up now to view full answer, or log in if you already have an account!
Answer
To find fg(x), we first compute g(x) = ln x. Now substituting into f, we have:
Find the exact value of x for which f(2x + 3) = 6.
96%
101 rated
Only available for registered users.
Sign up now to view full answer, or log in if you already have an account!
Answer
Set f(2x + 3) to equal 6:
f(2x + 3) = e^{(2x + 3)} + 2 = 6.
Solving this gives:
e^{(2x + 3)} = 4
Taking natural log:
2x + 3 = ext{ln } 4
Thus,
2x = ext{ln } 4 - 3
x = rac{ ext{ln } 4 - 3}{2}.
Step 4
Find f^{-1}, the inverse function of f, stating its domain.
98%
120 rated
Only available for registered users.
Sign up now to view full answer, or log in if you already have an account!
Answer
To find the inverse function f^{-1}(y):
Start with y = f(x) = e^x + 2.
Rearranging gives:
y - 2 = e^x
Taking the natural log:
x = ext{ln}(y - 2)
Thus, the inverse function is f^{-1}(y) = ext{ln}(y - 2) ext{ for } y > 2.
Step 5
On the same axes sketch the curves with equation y = f(x) and y = f^{-1}(x), giving the coordinates of all the points where they cross the axes.
97%
117 rated
Only available for registered users.
Sign up now to view full answer, or log in if you already have an account!
Answer
To sketch:
The curve y = f(x) = e^x + 2 crosses the y-axis at (0, 3) because f(0) = e^0 + 2 = 3.
The curve y = f^{-1}(x) = ext{ln}(x - 2) crosses the x-axis at (2, 0) since f^{-1}(2) = ext{ln}(0) is undefined, but it approaches this point. The curves are symmetric along the line y = x.
