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 September 21st, 2011, 02:22 PM #1 Newbie   Joined: Sep 2011 Posts: 18 Thanks: 0 Rational function, exponential function, extrema Hi! Could someone explain to me, step by step, how to find maximum and minimum of a function like this: $f(x)= \frac{4x^2+2x+1}{x^2+x+1}$ ?
 September 21st, 2011, 04:41 PM #2 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,205 Thanks: 512 Math Focus: Calculus/ODEs Re: Rational function, exponential function, extrema We are given: $f(x)=\frac{4x^2+2x+1}{x^2+x+1}$ To find the extrema, we look for values where f'(x) = 0 or undefined: Using the quotient and power rules for differentiation, we obtain: $f'(x)=\frac{$$x^2+x+1$$$$8x+2$$-$$4x^2+2x+1$$$$2x+1$$}{$$x^2+x+1$$^2}$ We see that the denominator has no real roots, thus we need only simplify the numerator and find its roots: $$$x^2+x+1$$$$8x+2$$-$$4x^2+2x+1$$$$2x+1$$=0$ $8x^3+2x^2+8x^2+2x+8x+2-$$8x^3+4x^2+4x^2+2x+2x+1$$=0$ $2x^2+6x+1=0$ Apply quadratic formula: $x=\frac{-6\pm\sqrt{36-4(2)(1)}}{2\cdot2}=\frac{-6\pm2\sqrt{7}}{4}=\frac{-3\pm\sqrt{7}}{2}$ These two roots will be where f(x) has its possible extrema. Dividing the number line at these critical values, and checking the sign of f'(x) in the resulting intervals shows: Interval $$$-\infty,-\frac{3+\sqrt{7}}{2}$$$: f'(-3) = (+)/(+) = + f(x) increasing on this interval. Interval $$$-\frac{3+\sqrt{7}}{2},-\frac{3-\sqrt{7}}{2}$$$ f'(-1) = (-)/(+) = - f(x) decreasing on this interval. Interval $$$-\frac{3-\sqrt{7}}{2},\infty$$$ f'(0) = (+)/(+) = + f(x) increasing on this interval. So, we find a local maximum at $$$-\frac{3+\sqrt{7}}{2},f\(-\frac{3+\sqrt{7}}{2}$$\)$ $f$$-\frac{3+\sqrt{7}}{2}$$=\frac{4$$-\frac{3+\sqrt{7}}{2}$$^2+2$$-\frac{3+\sqrt{7}}{2}$$+1}{$$-\frac{3+\sqrt{7}}{2}$$^2+$$-\frac{3+\sqrt{7}}{2}$$+1}=\frac{4$$9+6\sqrt{7}+7$$-4$$3+\sqrt{7}$$+4}{$$9+6\sqrt{7}+7$$-2$$3+\sqrt{7}$$+4}=$ $\frac{36+24\sqrt{7}+28-12-4\sqrt{7}+4}{9+6\sqrt{7}+7-6-2\sqrt{7}+4}=\frac{56+20\sqrt{7}}{14+4\sqrt{7}}=\f rac{4\sqrt{7}+10}{\sqrt{7}+2}=$ $\frac{2$$5+2\sqrt{7}$$}{2+\sqrt{7}}\cdot\frac{2-\sqrt{7}}{2-\sqrt{7}}=\frac{2$$10-5\sqrt{7}+4\sqrt{7}-14$$}{4-7}=\frac{2$$4+\sqrt{7}$$}{3}$ Thus, the maximum is at the point $$$-\frac{3+\sqrt{7}}{2},\frac{2\(4+\sqrt{7}$$}{3}\)$ Since $\lim_{x\to \pm\infty}f(x)=4$, we know this maximum is a global maximum. See if you can now find the xy coordinates of the minimum at the other critical value, and identify it as a global minimum. One can also use f''(x) to identify the type of extrema using the concavity of f(x). We may compute: $f'(x)=\frac{2x^2+6x+1}{$$x^2+x+1$$^2}$ $f''(x)=\frac{$$x^2+x+1$$^2$$4x+6$$-$$2x^2+6x+1$$$$2\(x^2+x+1$$$$2x+1$$\)}{$$x^2+x+1$$ ^4}=$ $\frac{2$$x^2+x+1$$(2x+3)-2$$2x^2+6x+1$$(2x+1)}{$$x^2+x+1$$^3}=$ $\frac{2$$2x^3+3x^2+2x^2+3x+2x+3-\(4x^3+2x^2+12x^2+6x+2x+1$$\)}{$$x^2+x+1$$^3}=$ $-\frac{2$$2x^3+9x^2+3x-2$$}{$$x^2+x+1$$^3}$ Using a root-finding technique, we find f''(x) has roots at: $x\approx-4.07123,-0.754374,0.325603$ As there are no roots of multiplicity, and f''(0) > 0, we conclude for f(x): Concave up on (-?,-4.07123) U (-0.754374,0.325603) Concave down on (-4.07123,-0.754374) U (0.325603,?) We find f''(x) is negative at the smaller critical value (?-2.8228, indicating a maximum. You can verify that f''(x) is positive at the other critical value (?-0.177124) indicating a minimum. This is a topic in elementary calculus, so I have moved it to a more appropriate sub-forum.
 September 22nd, 2011, 11:46 AM #3 Newbie   Joined: Sep 2011 Posts: 18 Thanks: 0 Re: Rational function, exponential function, extrema Thank you very much. I tried using it for x=2^n and it doesn't work because x must be positive then. Could you tell me what changes if x=a^n?
 September 22nd, 2011, 12:45 PM #4 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,205 Thanks: 512 Math Focus: Calculus/ODEs Re: Rational function, exponential function, extrema If $x=ka^n$ then we may implicitly differentiate with respect to n by first converting from exponential to logarithmic form: Note: $0 $\ln(x)=\ln(ka^n)=\ln(k)+n\ln(a)$ $\frac{1}{x}\cdot\frac{dx}{dn}=\ln(a)$ $\frac{dx}{dn}=x\ln(a)=ka^n\ln(a)$ Since $a\ne1$ there are no extrema. You only have: global minimum of 0 and global maximum of ±? (same sign as k).
 September 22nd, 2011, 01:28 PM #5 Newbie   Joined: Sep 2011 Posts: 18 Thanks: 0 Re: Rational function, exponential function, extrema Thanks.

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# maximum value of a rational function

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