Science:Math Exam Resources/Courses/MATH105/April 2010/Question 06

MATH105 April 2010

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[hide] Question 06
Let $f(x)=\lim _{n\to \infty }\left[\left\{{\frac {1}{3}}+{\frac {1}{2+\left(1+{\frac {x-1}{n}}\right)^{3}}}+{\frac {1}{2+\left(1+{\frac {2(x-1)}{n}}\right)^{3}}}+\ldots \right.\right.$ $\ldots +\left.\left.{\frac {1}{2+\left(1+{\frac {(n-1)(x-1)}{n}}\right)^{3}}}\right\}{\frac {x-1}{n}}\right]$ where x ≥ 0. Find the equation of the tangent line to the graph y = f(x) at x = 1.

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If you are stuck, check the hints below. Read the first one and consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it! If after a while you are still stuck, go for the next hint.

[show] Hint 1
Try to write the sum above in summation notations (using $\Sigma$ ). Do you recognize a pattern?

[show] Hint 2
f(x) is the limit of a left Riemann sum and can hence be represented as an integral. Try to find that integral expression.

[show] Hint 3
Recall that the general form of a (left) Riemann sum is $\int _{a}^{b}f(y)\,dy=\lim _{n\rightarrow \infty }\sum _{j=0}^{n-1}\Delta y\,f(y_{j}),$ where $y_{j}=a+j\Delta y$ and $\Delta y={\frac {b-a}{n}}$ .

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[show] Solution
Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. We can recognize this as the limit of a Riemann sum. Let's first put this into summation notation: $f(x)=\lim _{n\rightarrow \infty }\sum _{j=0}^{n-1}{\frac {1}{2+(1+j(x-1)/n)^{3}}}\cdot {\frac {x-1}{n}}$ Note that when $j=0$ this gives us the 1/3 term in the sum. We put the factor $(x-1)/n$ first to resemble the general form of a (left) Riemann sum as given in Hint 3. Ultimately, we aim to express $f(x)$ as a definite integral i.e. something of the form $\int _{a}^{b}g(s)ds=\lim _{n\rightarrow \infty }\sum _{j=0}^{n-1}g(s_{j})\cdot \Delta s$ We choose $s$ here for ease of notation and not overuse the $x$ symbol. For $x$ fixed, we can read off an interval spacing of $\Delta s={\frac {x-1}{n}}$ Let's rewrite the sum with $\Delta s$ in place of $(x-1)/n$ : $f(x)=\lim _{n\rightarrow \infty }\sum _{j=0}^{n-1}{\frac {1}{2+(1+j\Delta s)^{3}}}\cdot \Delta s$ . This is suggestive of a left-endpoint Riemann sum, since we start with $j=0$ and end at $j=n-1$ . In that case, we can take our subinterval endpoints to be $s_{j}=1+j\Delta s$ (recall that when we do a Riemann sum, we have subintervals with endpoints that look like $s_{j}=a+j\Delta s$ and terms of the form $1+j\Delta s$ are of this form). Let's rewrite this as $f(x)=\lim _{n\rightarrow \infty }\sum _{j=0}^{n-1}{\frac {1}{2+s_{j}^{3}}}\cdot \Delta s$ From here, we read off $g(s)={\frac {1}{2+s^{3}}}$ Also, since $s_{0}=1+0\Delta s=1$ is the lower bound and $s_{n}=1+n\Delta s=1+n(x-1)/n=x$ is the upper bound, our final result is that $f(x)=\int _{1}^{x}{\frac {1}{2+s^{3}}}ds$ Finally we can compute the equation of the tangent line to $f(x)$ at $x=1$ . All we need are $f(1)$ and $f'(1)$ . $f(1)=\int _{1}^{1}g(s)ds=0$ since we are integrating over a range of zero. And by the Fundamental Theorem of Calculus, $f'(x)={\frac {1}{2+x^{3}}}$ so $f'(1)={\frac {1}{3}}$ Hence the equation of the tangent line is $y-0={\frac {1}{3}}(x-1)$ or simply $y={\frac {1}{3}}(x-1)$