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junior college 1 | H2 Maths
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Hello! How do i solve exactly f inverse x =0? Just that part will do, no need the whole qns, thank you! :)
Date Posted: 4 years ago
Views: 240
Remember that a point on the function f(x) with coordinates (a,b) would have a corresponding point (b,a) on its inverse f-¹(x) when reflected along the line of y = x
For example, for the function y = 2x, the point (2,4) would have a corresponding point (4,2) on the inverse y = ½x
For example, for the function y = 2x, the point (2,4) would have a corresponding point (4,2) on the inverse y = ½x
So when asked to find f-¹(x) = 0, it is simply asking for the y coordinate of f-¹(x) to be equated to 0.
This corresponds to equating the x coordinate of the original function f(x) to 0
So when x = 0 , f(x) = 0³ + 0 - 7 = -7
This means x = -7 when f-¹(x) = 0
This corresponds to equating the x coordinate of the original function f(x) to 0
So when x = 0 , f(x) = 0³ + 0 - 7 = -7
This means x = -7 when f-¹(x) = 0
iii)
The intersection point of f and f-¹ is on the line y = x
So let y = f(x)= x³ + x - 7
Since y = x at the intersection point,
x = x³ + x - 7
x³ - 7 = 0
x³ = 7
x = ³√7
The intersection point of f and f-¹ is on the line y = x
So let y = f(x)= x³ + x - 7
Since y = x at the intersection point,
x = x³ + x - 7
x³ - 7 = 0
x³ = 7
x = ³√7
Ohh i see thank you so much for the explanation!! :)
Welcome
For part (ii), I do the following.
The fact that f-¹(x) = 0 (yes, I cannot find the operation for the power, so I copied the power of -1 from J) means that
f(0) = x
[If y = f(x), then f-¹(y) = x]
But f(0) = 0^3 + 0 - 7 = -7
So, -7 = x.
The fact that f-¹(x) = 0 (yes, I cannot find the operation for the power, so I copied the power of -1 from J) means that
f(0) = x
[If y = f(x), then f-¹(y) = x]
But f(0) = 0^3 + 0 - 7 = -7
So, -7 = x.
Another way to present :
Recall that ff-¹(x) = x and f-¹f(x) = x
So when f-¹(x) = 0,
x = ff-¹(x) = f(0) = 0³ + 0 - 7 = 0
Recall that ff-¹(x) = x and f-¹f(x) = x
So when f-¹(x) = 0,
x = ff-¹(x) = f(0) = 0³ + 0 - 7 = 0
tysm for the both of your help!! :D
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Good evening Candice! Here are my workings for parts ii and iii.
Date Posted:
4 years ago
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Good evening Candice! Here are my workings for part i using an “otherwise“ approach, where I do not fully plot the graph on a graphing calculator, but I use my knowledge of the basic x^3 graph and differentiation (to prove the absence of any stationary point) to solve this question.
Date Posted:
4 years ago
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