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# Tutor profile: Alex A.

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Alex A.
Tutor Coordinator at university for 10+ years.
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## Questions

### Subject:Pre-Calculus

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Question:

If $$f(x) = x^{2} + 3x - 1$$ and $$g(x) = 2x^{2} + 1$$ find $$(f \circ g)(x)$$.

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Alex A.

To evaluate $$(f \circ g)(x)$$, we find that: $$(f \circ g)(x) = f(g(x)) = f(2x^{2} + 1)$$ Substituting $$(2x^2 + 1)$$ into our composition gives us: $$f(2x^{2} + 1) = (2x^{2} + 1)^{2} + 3(2x^{2} + 1) - 1$$ Using $$(a + b)^{2} = a^{2} + 2ab +b^{2}$$ and the distributive law we get: $$f(2x^{2} + 1) = (2x^{2} + 1)^{2} + 3(2x^{2} + 1) - 1$$ = $$4x^{4} + 4x^{2} + 1 + 6x^{2} + 3 + 1$$ = $$4x^{4} + 10x^{2} +3$$

### Subject:Trigonometry

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Question:

Solve the following equation and list the solutions which lie in the interval $$[0,2\pi)$$. $$\\2\sin^{3}(x) = \sin^{2}(x)$$

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Alex A.

Set equation equal to zero by moving all terms to the left side. $$\\2\sin^{3}(x) - \sin^{2}(x) = 0$$ Factor out the common term of $$\sin^{2}(x)$$. $$\\\sin^{2}(x)(2\sin(x)-1) = 0$$ (Using zero factor property, set each equal to zero to solve.) $$\\\sin^{2}(x)=0$$ and $$2\sin(x)-1 = 0$$ Using the unit circle, we find that the angles that solve $$\\\sin^{2}(x)=0$$ are: $$0$$ and $$\pi$$ Isolating for $$\sin(x)$$ in the equation: $$\\2\sin(x)-1 = 0$$ (leads us to) $$\sin(x) = \frac{1}{2}$$ (Using unit circle we find that angles are: $$\frac{\pi}{6}$$ and $$\frac{5\pi}{6}$$

### Subject:Calculus

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Question:

Find the slope of the tangent line at the point P = (1,1) on the graph of $$e^{x-y}$$ on the graph of $$e^{x-y} = 2x^{2} - y^{2}$$.

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Alex A.

$$\frac{d}{dx}e^{x-y}=\frac{d}{dx}(2x^{2}-y{2})$$ (Start by taking the derivative with respect to x.) $$\\e^{x-y}(1-y^{'})=4x-2yy^{'}$$ (Apply the chain rule to $$e^{x-y}$$) $$\\e^{x-y}-e^{x-y}y{'}=4x-2yy^{'}$$ $$\\(2y-e^{x-y})y{'}=4x-e^{x-y}$$ (Place all the $$y^{'}$$ terms on the left side.) $$\\y{'}=\frac{4x-e^{x-y}}{2y-e^{x-y}}$$ $$\\\frac{dy}{dx}\mid_{(1,1)}=\frac{4(1)-e^{1-1}}{2(1)-e^{1-1}}=\frac{4-1}{2-1}=3$$ (Evaluate $$y{'}$$ at P(1,1).)

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