# Tutor profile: Krystyna I.

## Questions

### Subject: Geometry

Define the equation of the tangent to parabola $$y = x^2 $$ parallel to the line $$y = x$$.

The equation of tangent to the line $$y = f(x)$$ at point $$x_0$$ in general case is given by $$y = f'(x_0)(x-x_0)+f(x_0)$$. We don't know, at which point we should have a tangent, however, we do know that it is parallel to the line $$y = x$$, from which we can immediatly say that $$f'(x_0) = 1$$. This is coming from the fact, that two lines $$ y = k_1x + b_1$$ and $$y = k_2x+b_2$$ if $$k_1 = k_2$$. Having defined that, we can find the point where we should compute the tangent: $$f'(x) = (x^2)' = 2x \implies 2x_0 = 1 \implies x_0 = \frac{1}{2}$$. The value of the function at this point is $$f(x_0) = \Big(\frac{1}{2}\Big)^2 = \frac{1}{4}$$. So, the final equation of the tangent to the parabola is given by $$y = \Big(x-\frac{1}{2}\Big) + \frac{1}{4} = x - \frac{1}{4}$$.

### Subject: Calculus

Compute the integral $$\int\frac{e^{2x}}{1+e^x}dx$$

The nominator is $$e^{2x} = e^xe^x$$. One of the exponents we can put inside the $$dx$$ by integrating it. Since the integral of the exponential function is again the exponential function, our initial integral reads: $$\int\frac{e^x}{1+e^x}d(e^x)$$. Now we can replace a variable as follows $$e^x = u$$, which leads to $$\int\frac{u}{1+u}du$$. Let us consider the expression inside the integral: $$\frac{u}{1+u} = \frac{1-1+u}{1+u} = \frac{(1+u)-1}{1+u} = 1-\frac{1}{1+u}$$, which means that we can re-write our integral as a sum of two: $$\int 1du - \int\frac{1}{1+u}du $$. The first integral will give us just $u$ as a result, while in the second one we can replace $$du$$ by $$d(u+1)$$ because the derivatives of these two functions are the same and they are equal to 1. So, we get $$u - \int\frac{1}{1+u}d(1+u) + c_1$$. The constant $$c_1$$ comes out after the integration of the first integral. The second integral is a standard one and it gives the logarithmic function as an outcome: $$u - \ln(1+u) + c_1 + c_2$$ = $$u - \ln(1+u) + c$$. We replaced $$c_1+c_2$$ by $$c$$ since the some of two constants its again a constant. Finally, going back to our initial variable $$x$$, we can write the answer: $$e^x - \ln(1+e^x) + c$$. You can check the answer by taking the derivative of the last expression: $$\Big(e^x-\ln(1+e^x)+c\Big)' = e^x-\frac{1}{1+e^x}e^x = \frac{e^x+e^{2x}-e^x}{1+e^x} = \frac{e^{2x}}{1+e^x}$$, which is exactly our initial function under the integration.

### Subject: Algebra

Find all the values of the parameter $$a$$ such that the equation $$\frac{6+a}{3+x}-\frac{6-a}{3-x} = \frac{6}{a}$$ has only one root.

First of all, we need to exclude the values for both unknown $$x$$ and the parameter $$a$$ which turn the denominator into zero. These values are $$(x \neq -3) \bigcup (x \neq 3)$$ and $$a\neq 0$$. Now let us simplify the equation, putting the left-hand side to one common denominator: $$\frac{18-6x+3a-ax-18-6x+3a+ax}{9-x^2} = \frac{6}{a}$$, which leads to equation $$\frac{6a-12x}{9-x^3} = \frac{6}{a}$$. Dividing both parts of the equation by 6, we can rewrite it in the following way $$\frac{a-2x}{9-x^2} = \frac{1}{a}$$, which finally will lead to the simple quadratic equation $$x^2-2ax+a^2-9 = 0$$. There are multiple ways to find the roots of the last equation, for example: $$(x-a)^2 - 9 = 0 \implies (x-a)^2-3^2 = 0 \implies (x-a-3)(x-a+3) = 0$$ $$x_1 = a-3, x_2 = a+3$$. Remember, that we have to find those values of the parameter $a$ which will give us only one root. Let's do the check $$x_1 = x_2 \implies a-3 = a+3$$, which has no solution. However, we have restrictions on the roots. By setting $$x = 3$$ we find that $$a_{1,2} = 0,6$$ and $$x = -3$$, $$a_{1,2} = 0,-6$$. The value $$a = 0$$ is out of our interests as we said ealier. Which means that for the values $$a = -6$$ and $$a = 6$$ the initial equation will have one root only.

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