January 10th, 2019, 01:34 AM  #1 
Member Joined: Feb 2018 From: Iran Posts: 55 Thanks: 3  Derivative and continuity Given the continuous function find a, b and c such that its graph has a tangent touching it at three points. 
January 10th, 2019, 05:48 AM  #2 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,690 Thanks: 2669 Math Focus: Mainly analysis and algebra 
The fact that it is continuous gives you $c$ and an equation for $a$ and $b$. For the third constraint, I would suppose that you find a tangent that touches both the first and the third part, and then determine a condition on $a$ and $b$ such that it touches the middle part. 
January 10th, 2019, 11:09 AM  #3 
Banned Camp Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 126 
Left curve: y=x$\displaystyle ^{2}$+10x+8 is a vertical parabola with vertex at (5,17) and going through (2,2). Center curve: Any parabola ax$\displaystyle ^{2}$+bx+c going through (2,2) and (0,0) Right curve: Parabola sweeping up to right from (0,0) Any tangent line to left curve with slope greater than the one which goes through (0,0) will cut center curve and right curve. Line through (0,0): y=mx Slope of left curve: 2x+10 = m Point on line has to equal point on curve: y=(2x+10)x=x$\displaystyle ^{2}$+10x+8 $\displaystyle \rightarrow$ x=2$\displaystyle \sqrt{2}$ The tangent to x$\displaystyle ^{2}$+10x+8 between x=2$\displaystyle \sqrt{2}$ and x=2 will touch all three curves. EDIT (OP implies tangent can be anywhere) For center section, y' must be $\displaystyle \geq$ 0 to cut in three places. y'=2ax+b$\displaystyle \geq$0 From right boundary, c=0 From left boundary, b=2a4 So you have to satisfy a(1+x) $\displaystyle \geq$ 2 2<x<0 Last edited by skipjack; January 10th, 2019 at 12:06 PM. 
January 10th, 2019, 01:43 PM  #4 
Global Moderator Joined: Dec 2006 Posts: 21,026 Thanks: 2257  
January 10th, 2019, 10:40 PM  #5 
Banned Camp Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 126 
Forget EDIT portion of my previous post. The original is correct: Calculate point on lsft curve where tangent goes through origin. Any point on left curve to right of that point will have a steeper slope and thus cross ANY curve in the center portion and the right curve.  There are also three cutting points if center curve has horizantal tangent for y>0. Equation of center curve which satisfies BC's (2,8} and (0,0) $\displaystyle y=ax^{2}+(2a+4)x$ $\displaystyle y'=0 \rightarrow 2ax+2a+4=0 \rightarrow x=(1+\frac{2}{a})$ $\displaystyle y=a(1+\frac{2}{a})^{2}(2a+4)(1+\frac{2}{a})>0 \rightarrow a<3$ With $\displaystyle x=(1+\frac{2}{a})$ you also need $\displaystyle 2<x<0 \rightarrow a<2$ and $\displaystyle a>2$ Therefore the center curve $\displaystyle y=ax^{2}+(2a+4)x$ has a horizantal tangent at $\displaystyle x=(1+\frac{2}{a})$ which cuts the other two curves if 2<a<3. The most general analytic approach for center curve, without thinking about what geometry, is: Given center curve y=ax^{2}+(2a+4)x find the equation of a tangent line and require that it hits both the left and the right curves, ie, If y=mx+d is equation of tangent line through point (x_{1},y_{1}) on center curve, then y=x+x^2 and y=x^2+10x+8 must be solvable in their region of definition for appropriate x_{1}in the center region. Too much trouble unless you want to practice your algebra. (Had to drop Latex in this paragraph because for some reason it wasn't working.) My first post answered OP. The EDIT was rushed because I had to go out. This post replaces it. Last edited by zylo; January 10th, 2019 at 10:43 PM. 
January 11th, 2019, 05:13 AM  #6 
Banned Camp Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 126 
Various tangents with 3pt contact, You really don't need an image if you look for center curve with horizontal tangent for positive y (post #5) because left and right curves rise indefinitely. IMG_1683.jpg There is another interesting solution brought out by the picture above. If a is chosen so that y for the center curve (satisfying BC's) is positive, then for all x between 2 and the apex, there is a tangent that intersects all three curves. Last edited by skipjack; January 11th, 2019 at 06:46 PM. 
January 11th, 2019, 06:50 PM  #7 
Global Moderator Joined: Dec 2006 Posts: 21,026 Thanks: 2257 
You seem to have misunderstood the problem, as it refers to a tangent that touches the graph at three points, not a tangent that intersects the graph in three places.

January 12th, 2019, 05:31 AM  #8  
Banned Camp Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 126  Quote:
Actually easier than what I was doing, which required working with ranges of variables. Thanks for pointing that out. Sometimes you don't see the tree for the forest. My one concern is the red "Not secure" sign on address bar. It comes on whenever I respond. Back to the problem. You have a line y=mx+b and two conditions, tangency at left and right curves, to determine m and b. Then find tangent center curve using remaining parameter a. EDIT: Equation of center curve which satisfies BC's (2,8} and (0,0) $\displaystyle y=ax^{2}+(2a+4)x =mx+b$ $\displaystyle y'= 2ax+2a+4=m$ and solve for a. Two eqs in two unknowns: x and a. Last edited by zylo; January 12th, 2019 at 06:25 AM.  
January 12th, 2019, 11:36 AM  #9 
Global Moderator Joined: Dec 2006 Posts: 21,026 Thanks: 2257 
It would be a good idea to use a different letter for b and $b$.

January 12th, 2019, 12:36 PM  #10 
Senior Member Joined: May 2016 From: USA Posts: 1,310 Thanks: 552 
I pity the OP. This thread went completely off the rails. Archie started it perfectly. But I fear that the OP will get lost. $f(x) \text { is continuous } \implies \displaystyle \lim_{x \rightarrow \lambda} f(x) = f( \lambda ).$ $\therefore \displaystyle \left ( \lim_{x \rightarrow 2} ax^2 + bx + c \right ) = f(\ 2) = (\ 2)^2 + 10(\ 2) + 8 = \ 8$ $\displaystyle \text {and } \left ( \lim_{x \rightarrow 0} ax^2 + bx + c \right ) = f(0) =(0)^2 + 2(0) = 0.$ $\displaystyle 0 = \left ( \lim_{x \rightarrow 0} ax^2 + bx + c \right ) = \left ( a * \lim_{a \rightarrow 0} x^2 \right) + \left ( \lim_{b * \rightarrow 0} x \right ) + \left ( \lim_{x \rightarrow 0} c \right ) \implies$ $0 = 0 + 0 + c \implies c = 0 \implies f(x) = ax^2 + bx \text { if } \ 2 < x < 0.$ $\displaystyle \left ( \lim_{x \rightarrow 2} ax^2 + bx \right ) = \ 8 \implies a(\ 2)^2 + b(\ 2) =  8 \implies$ $4a  2b = \ 8 \implies 2a  b = \ 4.$ But f(x) consists of 3 segments, each apparently quadratic in form. The tangent of a true quadratic intersects that quadratic only at the point of tangency. Because there is a single line that is tangent to f(x) in 3 distinct places, there must be one point of tangency in each segment. Let p and q be the points of tangency when x $\le$ 2 and x $\ge$ 0 respectively. $\therefore f(p) = p^2 + 10p + 8,\ f(q) = q^2 + 2q,$ $f'(p) = 2p + 10, \text { and } f'(q) = 2q + 2.$ But f(p) and f(q) lie on the same line so f'(p) and f'(q) must be equal. $2p + 10 = 2q + 2 \implies q = p + 4.$ $\therefore (q  p) = 4 \text { and } q^2 = p^2 + 8p + 16.$ The equation of the line joining the 2 points of tangency is $\dfrac{y  (q^2 + 2q)}{x  q} = \dfrac{(q^2 + 2q)  (p^2 + 10p + \text {8})}{q  p} \implies$ $\dfrac{y  (p^2 + 8p + 16 + 2p + 8}{x  (p + 4)} = \dfrac{p^2 + 8p + 16 + 2p + 8  p^2  10p  8}{(p + 4)  p} \implies$ $\dfrac{y  p^2  10p  24}{x  p  4} = \dfrac{16}{4} = 4 \implies$ $y = 4x  4p  16 + p^2 + 10p + 24 = 4x + p^2 + 6p + 8.$ So the slope of the line is 4. Now we need Skipjack's contribution, which I did not see immediately whence it came. $f'(q) = 4 \implies 2q + 2 = 4 \implies q = 1 \implies$ $f(q) = 3, \ p = 1  4 = \ 3, \text { and } f(p) = (\ 3)^2 + 10(\ 3) + 8 = \ 13.$ Also the line has the equation $y = 4x + (\ 3)^2 + 6(\ 3) + 8 = 4x  1.$ Back to the middle segment. We know that at the point of tangency $2ax + b = 4.$ $2ax + 2a + 4 = 4 \implies 2ax =  2a \implies x = \ 1 \text { or } a = 0.$ $a = 0 \implies b = 4 \implies f(x) = 4x \ne 4x  1 \implies a \ne 0.$ $\therefore x = \ 1.$ $x = \ 1 \implies 4x  1 = \ 5.$ $\therefore \ 5 = a(\ 1)^2 + b(1) = a  b \implies b = a + 5$ And earlier it was found that $2a  b =  4 \implies b = 2a + 4.$ $\therefore 2a + 4 = a + 5 \implies a = 1 \implies b = 6.$ So we get $x \le \ 2 \implies f(x) = x^2 + 10x + 8.$ $\ 2 < x < 0 \implies f(x) = x^2 + 6x.$ $0 \le x^2 + 2x.$ There is a common tangent $y = 4x  1$ at x =  3,  1, and 1. Let's check. $f(0) = 0 = 0^2 + 6(0).$ Continuous. $f(\ 2) =  8 = 4  12 = (\ 2)^2 + 6(\ 2).$ Continuous. $f(\ 3) = (\ 3)^2 + 10(\ 3) + 8 = 9  30 + 8 = \ 13 = 4(\ 3)  1.$ $f'(\ 3) = 2(\ 3) + 10 = 4.$ Good here. $f(\ 1) = (\ 1)^2 + 6(\ 1) =  5 = 4(\ 1)  1.$ $f'(\ 1) = 2(\ 1) + 6 = 4.$ Good here. $f(1) = 1^2 + 2(1) = 3 = 4(1)  1.$ $f'(1) = 2(1) + 2 = 4.$ Good here. 

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