1)If f(x) = x3+x+1 find d/dx (f-1(x))
there as 4 choices, A 3x2+1 B 1/X3+X+1 C 1/ 3X2+1 D 1/3y2+1 E NOTA
2). What is the instantaneous rate of change of f(x) = ln(csc2 x) at π/4
(a) 0 (b) 2 (c) - 2 (d) 1 (e) − ∞
3)Rewrite (3logx+1/2log(x-2))-log(x+4) as a single logarithm.
+118613 1)If f(x) = x3+x+1 find d/dx (f-1(x))
there as 4 choices, A 3x2+1 B 1/X3+X+1 C 1/ 3X2+1 D 1/3y2+1 E NOTA
I'm going for 1/(3y^2+1)
I will show you how i did it but I think I might get my hand slapped by a better mathematician. LOL
Even if my answer is right i will probably be told that my method suxs :)
That is ok, I want to be corrected if it is not correct or not presented properly :)
original function
y=x^3+x+1
y'=3x^2+1
No stationary points
gradient is always positive.
This means that the function is a 1 to 1 mapping and therefore the inverse function is too.
So the domain for the function and the inverse function are both all real x
and the range for the function and the inverse function are both all real y
here is the graphs https://www.desmos.com/calculator/m457r8jips
\(function\\ y=x^3+x+1\\ \frac{dy}{dx}=3x^2+1\\ inverse\; function\\ x=y^3+y+1\\ \frac{dx}{dy}=3y^2+1\\ \frac{dy}{dx}=\frac{1}{3y^2+1}\\ \)
+118613 3)Rewrite (3logx+1/2log(x-2))-log(x+4) as a single logarithm.
\((3logx+1/2log(x-2))-log(x+4) \\ =3logx+0.5log(x-2)-log(x+4) \\ =logx^3+log(x-2)^{0.5}-log(x+4) \\ =log \left(\frac{x^3(x-2)^{0.5}}{x+4}\right) \\ =log \left(\frac{x^3 \sqrt{x-2}}{x+4}\right) \\\)
+118613 2). What is the instantaneous rate of change of f(x) = ln(csc2 x) at π/4
(a) 0 (b) 2 (c) - 2 (d) 1 (e) − ∞
\(\boxed{y=ln(g(x))\\\frac{dy}{dx}=\frac{g'(x)}{g(x)}}\\ g(x)=cosec^2x\\ g(x)=(sinx)^{-2}\\ g'(x)=-2(sinx)^{-3}cosx\\ ....\\ f(x)=ln(cosec^2x)\\ f'(x)=\frac{-2(sinx)^{-3}cosx}{cosec^2x}\\ f'(x)=\frac{-2cosx*sin^2x}{(sinx)^{3}}\\ f'(x)=\frac{-2cosx}{sinx}\\ f'(x)=-2cotx\\ f'(\pi/4)=-2cot(\pi/4)\\ f'(\pi/4)=-2\)
