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# Let f be the real-value function defined by f(x)= sq root of(1+6x).?

Topic: Let f be the real-value function defined by f(x)= sq root of(1+6x).?
June 16, 2019 / By Anne
Question: ...Given the coordinates of the point on the graph of f where the tangent line is parallel to y= x+12... In the previous problem it told me to determine the tangent line at x=4... so the tangent line= y= 1/5(3x+ 13) at x=4.. but how would i do the problem above?

## Best Answers: Let f be the real-value function defined by f(x)= sq root of(1+6x).?

Wright | 4 days ago
given the curve y = f(x) = √(1 + 6x) find the coordinates of the point on the graph whose tangent line is parallel to the line y = x + 12 To solve this kind of problem, you have to know calculus. You need to take the derivative f'(x) of the function. The derivative means the slope of the curve(or the slope of the tangent line to a curve). Just read calculus books to know how to get the derivative of a function. In the meantime the derivative of f(x) is: f(x) = √(1 + 6x) f'(x) = 1/2 · 1/√(1 + 6x) · 6 f'(x) = 3/√(1 + 6x) Sinde f'(x) is the derivative, then it is the function of the slope of the tangent line to the curve at any point. since the tangent is to be parallel to y = x + 12, then they have equal slopes. Since the slope of y = x + 12 is 1, then the slope of the tangent line is also 1. 3/√(1 + 6x) = 1 solving for x, 3 = √(1 + 6x) √(1 + 6x) = 3 1 + 6x = 9 6x = 8 x = 4/3 When x = 4/3, then y or f(x) is f(4/3) = √[1 + 6(4/3)] f(4/3) = √(1 + 8) f(4/3) = √9 f(4/3) = 3 .·. the coordinates of the point is (4/3,3) ^_^
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Originally Answered: How do you tell if a quadratic will have real, non-real, or a double root?
Well, the easiest way to figure this out is to use a part of the quadratic formula: b² - 4ac For an equation to have real roots, the part in the square root of the quadratic equation (b² - 4ac) has to be greater than 0 or positive (because any positive number square rooted has two answers) For an equation to have non-real roots, the part in the square root of the quadratic equation (b² - 4ac) has to be less than 0 or negative (its non-real because you can't square a negative number) For an equation to have double roots, the part in the square root of the quadratic equation (b² - 4ac) has to be equal to 0 Ex. = x^2+x+1 (sub in the proper numbers into b² - 4ac) = (1)² - 4(1)(1) (solve) = 1 - 4 = -3 Since the number is negative or less than 0, that equation has non-real roots

Shaun
The line y = x + 12 has slope 1. So you try to find a point on the graph of f with slope 1. The derivative [1] ... f'(x) = 3/sqrt(1 + 6x) has slope 1 if [2] ... 3/sqrt(1 + 6x) = 1 [3] ... 1 + 6x = 9 [4] ... x = 4/3 The corresponding y-coordinate is [5] ... f(x) = sqrt(1 + 6x) = sqrt(9) = 3 So the point is (4/3, 3).
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Nidhogg
Take the first derivative of the function. From the given line, one can see that M, the slope, is equal to one. Set the first derivative equal to one and solve for x. The solution is x= 4/3.
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Kolby
Take the derivative of f(x) and then equate the slopes (the slope equals 1) and then determine the desired point.
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Imla
the best thing to do is start plugging in values to the equation and graph it out. this will start painting a picture for you and you will be able to understand it much better. if this doesnt help, let me know and i will help you further.
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Originally Answered: Can anyone give me a REAL LIFE problem about " quadratic functions, cubic functions, and square root functions?
Let's say that you give me one penny today, two pennies tomorrow, three pennies the next day, and so on. I want to know, first off, how much money I will have at the end of one year (365 days -- no leap day), and I want to know how long it will take for me to have \$1,000,000 at this rate. The answer to this is to count S(n) = 1 + 2 + ... + n. I notice that there are n terms and that they are bounded from above by n, so I notice that S(n) ≤ n² for all n. This means that I can probably work out a quadratic function A n² + Bn + C for S(n). So, I try to determine A, B, and C: S(n) = A n² + Bn + C. I first notice that C is very easy: the sum of the first 0 counting numbers is, well, zero. So: S(0) = C = 0. But what I really need is this information: "to get from the n-1'th number to the n'th number, you add n:" S(n) = S(n-1) + n. Between the two statements I've just made, this gives us everything we need: to get to S(1), you take S(0) = 0 and add 1. To get to S(2), you take S(1) = 1, and add 2. To get to S(3), you take S(2) = 3, and add 3. And so on. So we must force that C = 0, and we must force that this last equation holds well. To force it with our polynomial expression, we write: A n² + Bn = A (n − 1)² + B (n − 1) + n Unfactoring: A n² + Bn = A (n² − 2 n + 1) + B (n − 1) + n A n² + Bn = A n² − 2 A n + A + B n − B + n The A n² + Bn can now be cancelled from both sides: 0 = − 2 A n + A − B + n. 0 = n (1 − 2 A) + A − B. To force that this is equal to 0 for all n, we must have that (1 − 2 A) = 0, and A = B. The first really says that A = ½, so: S(n) = ½ n² + ½ n. Okay! now what about S(365), the number of pennies you'll have given me after one year? You will be giving me \$3.65 on the last day, of course, but how much is the sum total you have given me? It will be: S(365) = (365 * 365 + 365)/2 = 66795 ... which is \$667.95 . This should not surprise you -- you have been giving me over \$1 for the last 260 days, so the number *should* be several hundreds of dollars. Now we want to know when I will have 100 million pennies total. This requires that we solve this equation: S(n) = M M = ½ n² + ½ n 2M = n² + n. To figure out what n is, I observe that: (n + ½)² = n² + n + ¼ And thus that n² + n = (n + ½)² − ¼. This may not look important, but it is: 2 M = (n + ½)² − ¼ (n + ½)² = 2 M + ¼ We can now use the square root function: n + ½ = √( 2 M + ¼ ). Okay, so to get 100 million pennies, I will want to calculate on my calculator: n = sqrt(200 000 000.25) - 0.5 = 14 141.635... This is a lot of days! But it is not *too* many. You would give me my hundred-millionth penny on day 14,142, thus completing one million dollars. It would take you a little under thirty-nine years of daily payments in order to pay it all. It might surprise you that such quadratic (n²) payments can rise so large as a million dollars over 40 years, from just a single penny. But actually, in the real world, debts grow by an asymptotically faster function, which grows like kⁿ. They choose a very small k -- perhaps k = 1.001, so that if you owe \$10 right now you should be paying one penny per day, or \$3.65 per year -- a 36.5% "interest rate". But if you don't pay those little daily pennies for ten years or so, that \$10 will have accumulated \$384 debt. Leave it another ten years, and it will cost you nearly \$15,000. It grows /exponentially/ -- if you owe twice as much, then your debt grows twice as fast.

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