Die diagram hieronder toon die grafieke van $g(x) = \frac{2}{x + p} + q$ en $f(x) = \log_{g} x$ - NSC Mathematics - Question 5 - 2017 - Paper 1
Question 5
Die diagram hieronder toon die grafieke van $g(x) = \frac{2}{x + p} + q$ en $f(x) = \log_{g} x$.
- $y = -1$ is die horisontale asymptoot van $g$.
- $B(1; 0)$ is die... show full transcript
Worked Solution & Example Answer:Die diagram hieronder toon die grafieke van $g(x) = \frac{2}{x + p} + q$ en $f(x) = \log_{g} x$ - NSC Mathematics - Question 5 - 2017 - Paper 1
Step 1
Skryf die waardeverzameling van g neer.
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
Die waardeverzameling van g is: y∈R;y≤−1.
Step 2
Bepaal die vergelyking van g.
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
Die vergelyking van g is: g(x)=x−22−1.
Step 3
Bereken die waarde van t.
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
Aan die hand van die grafiek, waar A(t;1) intersect met f, kan ons die waarde van t as t=3 bepaal.
Step 4
Skryf neer die vergelyking van f^{-1}(x), die inverse van f, in die vorm y = ...
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
Die inverse van f word gegee as: y=3x.
Step 5
Vir watter waardes van x sal f^{-1}(x) < 3?
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
Die waardes van x waarvoor f−1(x)<3 is wanneer x<1.
Step 6
Bepaal die snypunt van die grafiek van f en die simmetrie-as van g wat 'n negatiewe gradient het.
97%
121 rated
Only available for registered users.
Sign up now to view full answer, or log in if you already have an account!
Answer
Die simmetrie-as van g is gegee deur die vergelyking y=−x+1. Die snypunt met die grafiek van f gebeur by B(1;0).
