Which one of the following functions has a horizontal tangent at \((0, 0)\)?
- \(y=x^{-\frac{1}{3}}\)
- \(y=x^{\frac{1}{3}}\)
- \(y=x^{\frac{2}{3}}\)
- \(y=x^{\frac{4}{3}}\)
- \(y=x^{\frac{3}{4}}\)
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Which one of the following functions has a horizontal tangent at \((0, 0)\)?
\(D\)
\(\text{Index must be greater than 1 otherwise the gradient function }\)
\(\text{will not be defined at }x=0\)
\(\Rightarrow D\)
The graph of `f: [0, 1] -> R,\ f(x) = sqrt x (1 - x)` is shown below.
The edges of the right-angled triangle `ABC` are the line segments `AC` and `BC`, which are tangent to the graph of `f`, and the line segment `AB`, which is part of the horizontal axis, as shown below.
Let `theta` be the angle that `AC` makes with the positive direction of the horizontal axis, where `45^@ <= theta < 90^@`.
a. | `text(Area)` | `= int_0^1 (sqrt x – x sqrt x)\ dx` |
`= int_0^1 (x^(1/2) – x^(3/2))\ dx` | ||
`= [2/3 x^(3/2) – 2/5 x^(5/2)]_0^1` | ||
`= (2/3 – 2/5) – (0 – 0)` | ||
`= 10/15 – 6/15` | ||
`= 4/15\ text(units)^2` |
b. | `f (x)` | `= x^(1/2) – x^(3/2)` |
`f prime (x)` | `= 1/2 x^(-1/2) – 3/2 x^(1/2)` | |
`= 1/(2 sqrt x) – (3 sqrt x)/2` | ||
`= (1 – 3x)/(2 sqrt x)\ \ text(.. as required.)` |
c. `m_(AC) = tan 45^@=1`
`=> m_(BC) = -1\ \ (m_text(BC) _|_ m_(AC))`
`text(At point of tangency of)\ BC,\ f prime(x) = -1`
`(1 – 3x)/(2 sqrt x)` | `=-1` |
`1-3x` | `=-2sqrtx` |
`3x-2sqrt x-1` | `=0` |
`text(Let)\ \ a=sqrtx,`
`3a^2-2a-1` | `=0` |
`(3a+1)(a-1)` | `=0` |
`a=1 or -1/3` |
`:. sqrt x` | `=1` | `or` | `sqrt x=- 1/3\ \ text{(no solution)}` |
`x` | `=1` |
`f(1)=sqrt1(1-1)=0\ \ =>B(1,0)`
`:.\ text(Equation of)\ \ BC, \ m=-1, text{through (1,0) is:}`
`y-0` | `=-1(x-1)` |
`y` | `=-x+1` |
d. `text(Find Equation)\ AC:`
`m_(AC) =1`
`text(At point of tangency of)\ AC,\ f prime(x) = 1`
`(1 – 3x)/(2 sqrt x)` | `=1` |
`1-3x` | `=2sqrtx` |
`3x+2sqrt x-1` | `=0` |
`(3 sqrtx-1)(sqrtx+1)` | `=0` |
`:. sqrt x` | `=1/3` | `or` | `sqrt x=- 1\ \ text{(no solution)}` |
`x` | `=1/9` |
`f(1/9)=sqrt(1/9)(1-1/9)=1/3 xx 8/9 = 8/27\ \ =>P(1/9,8/27)`
`:.\ text(Equation of)\ AC, m=1, text(through)\ \ P\ \ text(is):`
`y – 8/27` | `= 1 (x – 1/9)` |
`y` | `= x + 5/27` |
`C\ text(is at intersection of)\ AC and CB:`
`-x + 1` | `= x + 5/27` |
`2x` | `= 22/27` |
`:. x` | `= 11/27` |
`y` | `= -11/27 + 1 = 16/27` |
`:. C (11/27, 16/27)`
Let `f: (−∞,1/2] -> R`, where `f(x) = sqrt(1 - 2x)`.
a. | `f(x)` | `= (1 – 2x)^(1/2)` |
`f′(x)` | `= 1/2(1 – 2x)^(−1/2) (−2)qquadtext([Using Chain Rule])` | |
`= (−1)/(sqrt(1 – 2x))` |
b. | `tan theta` | `= f′(−1)` |
`= (−1)/(sqrt(1 – 2(−1)))` | ||
`= (−1)/(sqrt3)` |
`text(S)text(ince)\ theta ∈ [0,pi],`
`=> theta = (5pi)/6`
A normal to the graph of `y = sqrt x` has equation `y = – 4x + a`, where `a` is a real constant. Find the value of `a.` (4 marks)
`a = 18`
`text(Normal) = text(line) _|_ text(to tangent)`
`m_{text(norm)}` | `=-4` |
`:. m_tan` | `=1/4` |
`y` | `= sqrt x` |
`dy/dx` | `=1/2 x^(-1/2)` |
`text(Find)\ x\ text(when:)`
`1/(2 sqrt x)` | `=1/4` |
`sqrt x` | `=2` |
`x` | `=4` |
`:.\ text(Point of tangency is)\ \ (4, 2)`
`text(Find equation of normal:)`
`y – y_1` | `= m (x – x_1)` |
`y – 2` | `= -4 (x – 4)` |
`:. y` | `= -4x + 18` |
`:. a` | `= 18` |
The normal to the graph of `y = sqrt (b - x^2)` has a gradient of 3 when `x = 1.`
The value of `b` is
A. `-10/9`
B. `10/9`
C. `4`
D. `10`
E. `11`
`D`
`y` | `=sqrt (b – x^2)` |
`dy/dx` | `=(-2x)/(2 sqrt(b-x^2))` |
`text(When)\ \ x=1,`
`dy/dx` | `=(-1)/sqrt(b-1)` |
`text(S)text(ince)\ \ m_(norm) xx m_(tan) = -1,`
`dy/dx=- 1/3`
`(-1)/sqrt(b-1)` | `=- 1/3` |
`sqrt(b-1)` | `=3` |
`b-1` | `=9` |
`b` | `=10` |
`=> D`