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

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?

Answer 1

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)

^_^

Answer 2

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).

Answer 3

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.

Answer 4

Take the derivative of f(x) and then equate the slopes (the slope equals 1) and then determine the desired point.

Answer 5

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.