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Tutor profile: Geoffrey L.

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Geoffrey L.
High School Mathematics Teacher for 4 years
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Questions

Subject:Trigonometry

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Question:

Solve the equation $$\sin (x)=\frac{1}{2}$$ for $$-\pi\le x\le\pi$$.

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Geoffrey L.

The solution to $$\sin(x)=\frac{1}{2}$$ for $$0\le x\le \frac{\pi}{2}$$ is $$x=\frac{\pi}{6}$$. This result may be obtained by observing the ratios of sides in a triangle formed from side lengths 1, 2 and $$\sqrt{3}$$ (half of an equilateral triangle). It may also be obtained by simply typing $$\sin^{-1}(\frac{1}{2})$$ into a calculator. This solution is not the only value of $$x$$ that produces the correct value. As the sine function repeats itself we need to search for more solutions within the given range. $$\sin x$$ is also positive for $$\frac{\pi}{2}\le x\le\pi$$ and we have the result $$\sin(x)=\sin(\pi-x)$$. This rule gives us the other solution: $$\pi-\frac{\pi}{6}=\frac{5\pi}{6}$$ Thus the two solutions to the original equations are $$x=\frac{\pi}{6}$$ and $$x=\frac{5\pi}{6}$$

Subject:Calculus

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Question:

Differentiate the following expression: $$y=\ln (x^2)$$

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Geoffrey L.

This expression is an example of a function ($$\ln$$) of a function ($$x^2$$). As such we must apply the "chain rule" (otherwise known as the function of a function rule) to differentiate it. let $$u=x^2$$ and differentiating this expression gives $$\frac{du}{dx}=2x$$. Returning to the original expression and substituting $$u$$ for $$x^2$$ we get $$y=\ln u$$. Differentiating this expression gives $$\frac{dy}{du}=\frac{1}{u}$$. The chain rule says $$\frac{dy}{dx}=\frac{dy}{du}\times\frac{du}{dx}$$. Applying this rule we get the following result: $$\frac{dy}{dx}=\frac {1}{u}\times 2x$$ $$\frac{dy}{dx}=\frac{1}{x^2}\times 2x$$ $$\frac{dy}{dx}=\frac{2x}{x^2}$$ $$\frac{dy}{dx}=\frac{2}{x}$$

Subject:Algebra

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Question:

Solve the following equation: $\frac{x}{3}+5=2$

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Geoffrey L.

Step 1: Subtract "5" from both sides of the equation: $\frac{x}{3}=-3$ Step 2: Multiply both sides by "3" and you arrive at the solution: $x=-9$

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