Let \(f(x)=\log_{e}x\), where \(x>0\) and \(g(x)=\sqrt{1-x}\), where \(x<1\).
The domain of the derivative of \((f\circ g)(x)\) is
- \(x\in R\)
- \(x\in (-\infty, 1]\)
- \(x\in (-\infty, 1)\)
- \(x\in (0, \infty)\)
- \(x\in (0, 1)\)
Aussie Maths & Science Teachers: Save your time with SmarterEd
Let \(f(x)=\log_{e}x\), where \(x>0\) and \(g(x)=\sqrt{1-x}\), where \(x<1\).
The domain of the derivative of \((f\circ g)(x)\) is
\(C\)
\(\text{Given }f(x)=\log_{e}x\) | \(\ \ \text{and}\ g(x)=\sqrt{1-x}\) |
\(\text{then }(f\circ g)(x)\) | \(=\log_{e}\sqrt{1-x}\) |
\(=\dfrac{1}{2}\log_{e}(1-x)\) | |
\((f\circ g)'(x)\) | \(=\dfrac{1}{2}\times\dfrac{-1}{1-x}\) |
\(=\dfrac{1}{2(x-1)}\ \text{ where}\ x<1\) |
\(\Rightarrow C\)
Let `f` and `g` be functions such that `f(-1)=4, \ f(2)=5, \ g(-1)=2, \ g(2)=7` and `g(4)=6`.
The value of `g(f(-1))` is
`D`
`f(-1)=4`
`g(f(-1)) = g(4) = 6`
`=>D`
Let `g(x) = x + 2` and `f(x) = x^2 - 4`
If `h` is the composite function given by `h : [–5, –1) to R, h(x) = f(g(x))`, then the range of `h` is
`E`
`h(x)` | `= (x + 2)^2 – 4` |
`= x^2 + 4x` |
`text{By CAS, graph} \ \ y = x^2 + 4x , \ x∈ [–5, –1)`
`text{Min at} \ (–2, –4)`
`text{Max at} \ (–5, 5)`
`:. \ text{Range of} \ h(x) ∈ [–4, 5]`
`=> E`
Let `f(x) = -x^2 + x + 4` and `g(x) = x^2 - 2`.
a. `f(3)` | `= -3^2 + 3 + 4` |
`= -2` |
`g(f(3))` | `= g(-2)` |
`= (-2)^2 – 2` | |
`= 2` |
b. `f(g(x))` | `= -(x^2 – 2)^2 + (x^2 – 2) + 4` |
`= -(x^4 – 4x^2 + 4) + x^2 + 2` | |
`= -x^4 + 5x^2 – 2` |
Let `f` and `g` be two functions such that `f(x) = 2x` and `g(x + 2) = 3x + 1`.
The function `f (g(x))` is
`D`
`g(x + 2)` | `= 3x + 1` |
`g((x – 2) + 2)` | `= 3(x – 2) + 1` |
`g(x)` | `= 3x – 5` |
`f(x)` | `= 2x` |
`f(g(x))` | `= 2(3x – 5)` |
`= 6x – 10` |
`=> D`
Let `f` and `g` be functions such that `f (2) = 5`, `f (3) = 4`, `g(2) = 5`, `g(3) = 2` and `g(4) = 1`.
The value of `f (g(3))` is
`D`
`f(g(3))` | `=f(2)` |
`=5` |
`=> D`
Let `f: [0, oo) -> R,\ f(x) = sqrt(x + 1)`.
a. `text(Sketch of)\ \ f(x):`
`:.\ text(Range)\ \ (f) = [1, oo)`
b.i. `text(Sketch)\ \ g(x) = (x + 1) (x + 3)`
`text(Domain of)\ \ f(x)=[0,oo)`
`text(Find domain of)\ \ g(x)\ \ text(such that Range)\ (g) = [0, oo)`
`text(Graphically, this occurs when)\ \ g(x)\ \ text(has domain:)`
`=> x ∈ (–oo, –3] ∪ [–1,oo)`
`:. c = -3`
b.ii. `text(Range)\ g(x) = [0, oo) = text(Domain)\ \ f(x)`
`:.\ text(Range)\ \ f(g(x)) = [1, oo)`
c. `text(Range)\ h(x) = [3, oo)`
`f(3)` | `= sqrt (3 + 1)` |
`= sqrt 4` | |
`= 2` |
`:.\ text(Range)\ \ f(h(x)) = [2, oo)`
Let `g(x) = x^2 + 2x - 3 and f(x) = e^(2x + 3).`
Then `f(g(x))` is given by
`D`
`text(Solution 1)`
`text(Define)\ \ f(x) and g(x)\ \ text(on CAS)`
`f(g(x)) = e^(2x^2 + 4x – 3)`
`=> D`
`text(Solution 2)`
`f(g(x))` | `=e^(2 xx (x^2 + 2x – 3)+3)` |
`= e^(2x^2 + 4x – 3)` |
`=>D`
Let `f : (0, ∞) → R`, where `f(x) = log_e(x)` and `g: R → R`, where `g (x) = x^2 + 1`.
a.i. | `h(x)` | `= f(x^2 + 1)` |
`= log_e(x^2 + 1)` |
a.ii. `text(Domain)\ (h) =\ text(Domain)\ (g) = R`
`text(For)\ x ∈ R` | `-> x^2 + 1 >= 1` |
`-> log_e(x^2 + 1) >= 0` |
`:.\ text(Range)\ (h) = [0,∞)`
a.iii. | `text(LHS)` | `= h(x) + h(−x)` |
`= log_e(x^2 _ 1) + log_e((−x)^2 + 1)` | ||
`= log_e(x^2 + 1) + log_e(x^2 + 1)` | ||
`= 2log_e(x^2 + 1)` |
`text(RHS)` | `= f((x^2 + 1)^2)` |
`= 2log_e(x^2 + 1)` |
`:. h(x) + h(−x) = f((g(x))^2)\ \ text(… as required)`
a.iv. `text(Stationary points when)\ \ h′(x) = 0`
`text(Using Chain Rule:)`
`h′(x)` | `= (2x)/(x^2 + 1)` |
`:.\ text(S.P. when)\ \ x=0`
`text(Find nature using 1st derivative test:)`
`:.\ text{Minimum stationary point at (0, 0)}.`
b.i. `text(Let)\ \ y = k(x)`
`text(Inverse: swap)\ x ↔ y`
`x` | `= log_e(y^2 + 1)` |
`e^x` | `= y^2 + 1` |
`y^2` | `= e^x – 1` |
`y` | `= ±sqrt(e^x – 1)` |
`text(But range)\ \ (k^(−1)) =\ text(domain)\ (k)`
`:.k^(−1)(x) = −sqrt(e^x – 1)`
b.ii. `text(Range)\ (k^(−1)) =\ text(Domain)\ (k) = (−∞,0]`
`text(Domain)\ (k^(−1)) =\ text(Range)\ (k) = [0,∞)`
If `f(x) = 1/2e^(3x) and g(x) = log_e(2x) + 3` then `g (f(x))` is equal to
`D`
`text(Define)\ \ f(x)= 1/2e^(3x), \ g(x)= log_e(2x) + 3`
`g(f(x))` | `= log_e(2 xx 1/2e^(3x)) + 3` |
`=log_e e^(3x) + 3` | |
`=3x + 3` | |
`= 3 (x + 1)` |
`=> D`
Let `f(x) = x^2 + 1 and g(x) = 2x + 1.` Write down the rule of `f(g(x)).` (1 mark)
`(2x + 1)^2 + 1`
`f (g(x))` | `=f(2x+1)` |
`= (2x + 1)^2 + 1` |
Let `f: R -> R,\ \ f(x) = e^(2x) - 1`.
a. `text(Let)\ \ y = f(x)`
`text(For Inverse, swap)\ x ↔ y`
`x` | `= e^(2y) – 1` |
`x + 1` | `= e^(2y)` |
`2y` | `= log_e(x + 1)` |
`y` | `= 1/2 log_e(x + 1)` |
`text(Domain)(f^(−1))` | `=\ text(Range)\ (f)` |
`= (−1,∞)` |
`:. f^(−1)(x) = 1/2log_e(x + 1),quadx ∈ (−1,∞)`
b. `f(f^(−1)(x)) = x`
`text(Domain is)\ \ (-1, oo)`
c. | `−f^(−1)(2x)` | `= −1/2 ln(2x + 1)` |
`:. f(−f^(−1)(2x))` | `= e^(-log_e(2x + 1)) – 1` | |
`=(2x+1)^-1 -1` | ||
`= 1/(2x + 1) – 1` | ||
`= (−2x)/(2x + 1)` |
If the function `f` has the rule `f(x) = sqrt (x^2 - 9)` and the function `g` has the rule `g(x) = x + 5`
a. | `f(g(x))` | `= sqrt {(x + 5)^2 – 9}` |
`= sqrt (x^2 + 10x + 16)` | ||
`= sqrt {(x + 2) (x + 8)}` |
`:. c = 2, d = 8 or c = 8, d = 2`
b. `text(Find)\ x\ text(such that:)`
`(x+2)(x+8) >= 0`
`(x + 2) (x + 8) >= 0\ \ text(when)`
`x <= -8 or x >= -2`
`:.\ text(Maximal domain is:)`
`x in (– oo, – 8] uu [– 2, oo)`