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# Tutor profile: Rabie R.

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Rabie R.
Experienced, Patient and Knowledgeable Math Tutor
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## Questions

### Subject:Pre-Calculus

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Question:

Consider the function $$f(x) = \ln(x-x^2)$$. Find the domain of $$f$$.

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Rabie R.

To find the domain, we have to find the values of $$x$$ for which $$f(x)$$ is defined. Since $$f(x)$$ is the log of some expression, the only restriction we have to consider is that what is inside the log has to be positive. So we get the inequality $(x-x^2 >0$) or $(x(1-x) >0$) The above inequality has two factors and the whole expression is positive if both factors are positive or both factors are negative. $$x$$ is positive when $$x>0$$ and negative when $$x<0$$. $$1-x$$ is positive when $$x<1$$ and negative when $$x>1$$. So, both factors are positive only when $$0<x<1$$. And both factors are never negative at the same time. Therefore, the domain of $$f(x)$$ is the open interval $$(0,1)$$.

### Subject:Calculus

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Question:

Find the equation of the tangent line to the function $(f(x) = \frac{1}{1+e^x}$) at the point where $$x=0$$.

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Rabie R.

First, find the slope $$m$$ of the tangent line at $$x=0$$. We know that $$m=f'(0)$$, so let us first calculate $$f'(x)$$: Use the quotient rule to find the derivative as follows: $(f'(x) = \dfrac{0(1+e^x) - 1(e^x)}{(1+e^x)^2} = \dfrac{-e^x}{(1+e^x)^2}$) So, $$f'(0) = \frac{-e^0}{(1+e^0)^2} = \frac{-1}{4}$$. Next, find the point on the graph of $$f(x)$$ when $$x=0$$. The point is $$(0,f(0))$$, or $$(0,\frac{1}{2})$$. Finally, the equation of the tangent line is $(y - \frac{1}{2} = \frac{-1}{4}(x-0)$) or $(y = -\frac{x}{4} +\frac{1}{2}$)

### Subject:Algebra

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Question:

Solve for $$x$$ in the following quadratic equation: $(2x^2 + 5x +3 = 0\;.$)

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Rabie R.

Using the Quadratic formula is the most straight forward way to solve a quadratic equation. First, calculate the discriminant $(\Delta = b^2-4ac = 5^2 - 4(2)(3) = 25 -24 =1 \;.$) Then, use the formula to get that $(x = \dfrac{-b\pm\sqrt{\Delta}}{2a} = \dfrac{-5 \pm \sqrt{1}}{2(2)} = \dfrac{-5\pm 1}{4}\;.$) So, the solutions to the equation are $$x=-1$$, and $$x=\frac{-3}{2}$$. Another way to solve the equation is to notice that we can factor it as follows: $(2x^2 + 5x +3 = 0\;.$) $((2x+3)(x+1) = 0\;.$) Then, we can immediately see that $$x=-1$$, or $$x=\frac{-3}{2}$$.

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