graph G(x)=[x]-1
A straight line, passing through the point (0,5) with a gradient of -3.
In A-sharp minor, every single note has a sharp. For the harmonic minor, the G♯ is raised to Gx (both ways) and for the melodic minor Fx and Gx is used on the way up but is reverted back to the key signature (normal F♯ and G♯ on the way down).
G of x= x2+1the G(0)= (0)2+10+1G(0)= 1
B# Major is not a real key signature. It's what is called an "imaginary key signature" - one that can be figured out theoretically, but isn't practical, and therefore not used. However, you can derive a scale from any note by going up the progression of T, T, ST, T, T, T, ST - so, the notes of B# major would be: B#, Cx, Dx, E#, Fx, Gx, Ax, B# (where "#" is a sharp and "x" is a double-sharp) If you are talking about just the chord of B# major then the same thing applies. It is an "imaginary chord" whose note would be: B#, Dx, Fx.
The values of x such as fgx= gfx is math. It comes down to finding the value of the letter X.
graph G(x)=[x]-1
What_is_the_area_bounded_by_the_graphs_of_fx_and_gx_where_fx_equals_xcubed_and_gx_equals_2x-xsquared
graph gx is the reflection of graph fx and then transformed 1 unit down
Since g(x) is known, it helps a lot to find f(x). f(g(x)) is a new function composed by substituting x in f with g(x). For example, if g(x) = 2x + 1 and f(g(x)) = 4x2+ 4x + 1 then you you recognize that this is the square of the binomial 2x + 1, so that f(g(x)) = (f o g)(x) = h(x) = (2x + 1)2, meaning that f(x) = x2. if you have a specific example, it will be nice, because there are different ways (based on observation and intuition) to decompose a function and write it as a composite of two other functions.
at first draw the graph of fx, then shift the graph along -ve x-axis 21 unit
A straight line, passing through the point (0,5) with a gradient of -3.
The composite function f of g is also expressed as f(g(x)). In this case, it would be 12(3x), or 36x.
Yes, the integral of gx dx is g integral x dx. In this case, g is unrelated to x, so it can be treated as constant and pulled outside of the integration.
4
g(x) = x + 3 Then f o g (x) = f(g(x)) = f(x + 3) = sqrt[(x+3) + 2] = sqrt(x + 5)
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