Let `f(x) = (ax + b)^5` and let `g` be the inverse function of `f`.
Given that `f(0) = 1`, what is the value of `g' (1)`
- `(5)/(a)`
- `1`
- `(1)/(5a)`
- `5a(a + 1)^4`
- `0`
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Let `f(x) = (ax + b)^5` and let `g` be the inverse function of `f`.
Given that `f(0) = 1`, what is the value of `g' (1)`
`C`
`f(x) = (ax + b)^5`
`text(Given) \ \ f(0) = 1\ \ =>\ \ b^5 = 1 \ => \ b = 1`
`f ‘(x) = 5a(ax + 1)^4`
`f(0) = 1 \ => \ g(1) = 0 \ \ text{(Inverse: swap} \ x ↔ y )`
`text(Gradient of) \ \ f(x) \ \ text(at) \ \ x =0 \ \ text(will be the reciprocal of the gradient of) \ \ g(x) \ \ text(at) \ \ x = 1`
`f ‘(0) = 5a`
`:. \ g'(1) = (1)/(5a)`
`=> \ C`
The function `f : D → R, \ f(x) = 5x^3 + 10x^2 + 1` will have an inverse function for
`E`
The function `f: B -> R` with rule `f(x) = 4x^3 + 3x^2 + 1` will have an inverse function for
`B`
The rule for `f` is `f(x) = x-1/2 x^2` for `x <= 1`. This function has an inverse, `f^(-1) (x)`.
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| a. |
| b. | `y = x-1/2 x^2,\ \ \ x <= 1` |
`text(For the inverse function, swap)\ \ x↔y,`
| `x` | `= y-1/2 y^2,\ \ \ y <= 1` |
| `2x` | `= 2y-y^2` |
| `y^2-2y + 2x` | `= 0` |
`text(Using quadratic formula,)`
| `y` | `= (2 +- sqrt( (-2)^2-4 * 1 * 2x) )/2` |
| `= (2 +- sqrt(4-8x))/2` | |
| `= (2 +- 2 sqrt(1-2x))/2` | |
| `= 1 +- sqrt (1-2x)` |
`:. y = 1-sqrt(1-2x), \ \ (y <= 1)`
| c. | `f^(-1) (3/8)` | `= 1-sqrt(1-2(3/8))` |
| `= 1-sqrt(1-6/8)` | ||
| `= 1-sqrt(1/4)` | ||
| `= 1-1/2` | ||
| `= 1/2` |
Let `f: R→R` where `f(x)= x^3-2`.
Evaluate `f^(-1)(25),` where `f^(-1)` is the inverse function of `f`. (2 marks)
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`3`
`text(Let)\ \ y = x^3-2`
`text(For inverse),\ \ x harr y`
| `x` | `= y^3-2` |
| `y^3` | `= x + 2` |
| `y` | `= (x + 2)^(1/3)` |
| `:. f^(-1)(25)` | `=(25+2)^(1/3)` |
| `=3` |
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a.i. `f^{prime}(x) = 12x^2 + 5`
a.ii. `text(S)text(ince)\ \ x^2>=0\ \ text(for all)\ x,`
| ` 12x^2` | `>= 0` |
| `12x^2 + 5` | `>= 5` |
| `f^{prime}(x)` | `>= 5\ \ text(for all)\ x` |
b.i. `p(x) = text(is a cubic)`
`:. m = 0, 1, 2`
`text{(Note: part a.ii shows that a cubic may have no SP’s.)}`
b.ii. `text(For)\ p^(−1)(x)\ text(to exist)`
`:. m = 0, 1`
c.i. `text(Let)\ y = q(x)`
`text(Inverse: swap)\ x ↔ y`
| `x` | `= 3-2y^3` |
| `y^3` | `= (3-x)/2` |
`:. q^(-1)(x) = root(3)((3-x)/2), \ x ∈ R`
c.ii. `text(Any function and its inverse intersect on)`
`text(the line)\ \ y=x.`
| `text(Solve:)\ \ 3-2x^3` | `= xqquadtext(for)\ x,` |
| `x` | `= 1` |
`:.\ text{Intersection at (1, 1)}`
| d.i. | `g^{prime}(x)` | `= 0` |
| `3x^2 + 4x + c` | `= 0` | |
| `Delta` | `= 0` | |
| `16-4(3c)` | `= 0` | |
| `:. c` | `= 4/3` |
d.ii. `text(Define)\ \ g(x) = x^3 + 2x^2 + 4/3x + k`
`text(Stationary point when)\ \ g^{prime}(x)=0`
`g^{prime}(x) = 3x^2+4x+4/3`
`text(Solve:)\ \ g^{prime}(x)=0\ \ text(for)\ x,`
`x = -2/3`
`text(Intersection of)\ g(x)\ text(and)\ g^(-1)(x)\ text(occurs on)\ \ y = x`
`text(Point of intersection is)\ (-2/3, -2/3)`
`text(Find)\ k:`
| `g(-2/3)` | `= -2/3\ text(for)\ k` |
| `:. k` | ` = -10/27` |
The function `f:\ (–oo, a] -> R` with rule `f(x) = x^3 - 3x^2 + 3` will have an inverse function provided
`A`
Consider the function `f: R -> R, \ f(x) = x(x - 4)` and the function
`g: [3/2,5) -> R, \ g(x) = x + 3`.
If the function `h = f + g`, then the domain of the inverse function of `h` is
`=> D`
Let `f: R^+ -> R` where `f(x) = 1/x^2.`
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| a. | `g(x)` | `= f(1/x^2)` |
| `= 1/((1/x^2)^2)` | ||
| `=1/(1/x^4)` | ||
| `:. g(x)` | `= x^4,\ \ x in R^+` |
b. `text(Let)\ \ y = f(x)`
`text(For inverse, swap)\ \ x harr y`
| `x` | `= y^4` |
| `y` | `= +- x^(1/4),` |
| `=x^(1/4)\ \ \ text{(Domain of}\ g(x)\ text(is)\ R^+ text{)}` |
| `:. g^-1 (16)` | `= 16^(1/4)` |
| `= 2` |
The rule for function `h` is `h(x) = 2x^3 + 1.` Find the rule for the inverse function `h^-1.` (2 marks)
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`h^-1 (x) = root 3 ((x-1)/2),\ \ x in R`
`h(x) = 2x^3+1,\ \ text(let)\ \ y = 2x^3+1,`
`text(For inverse, swap)\ x harr y`
| `x` | `= 2y^3 + 1` |
| `y^3` | `= (x-1)/2` |
| `y` | `= root 3 ((x-1)/2)` |
| `:. h^-1 (x)` | `= root 3 ((x-1)/2),\ \ x in R` |
The function `f: D -> R` with rule `f(x) = 2x^3 - 9x^2 - 168x` will have an inverse function for
`B`