Find a function f(x) = a + b sqrt(x) that is tangent to the line 2y - 3x = 5 at the point (1,4) AND Find a function of the form f(x)=c+dcoskx that is tangent to the line y=1 at the point (0,1) and tangent to the line y=x+(3/2) - (pi/4) at the point ([pi/4], [3/2)

To find a function that is tangent to a given line at a specific point, we need to determine the values of the coefficients in the function.

For the first problem, the given line is 2y - 3x = 5, and the point of tangency is (1, 4).

Step 1: Write the equation of the tangent line
Since the line is tangent to the given function, we also know that the slope of the tangent line is equal to the derivative of the function at the point of tangency.

The equation of the given line is 2y - 3x = 5.
Rearranging it to find the slope-intercept form: y = (3/2)x + 5/2.

So, the slope of the given line is m = 3/2.

Step 2: Write the equation of the tangent line in point-slope form
Using the point-slope form of a line equation: y - y1 = m(x - x1), we can plug in the point (1, 4) and the slope m = 3/2 to get:

y - 4 = (3/2)(x - 1)

Step 3: Express the equation in terms of f(x)
We need to find a function f(x) of the form a + b*sqrt(x) such that its derivative is equal to 3/2 at x = 1.

Differentiating f(x) = a + b*sqrt(x) with respect to x, we get:

f'(x) = (b/2)*(1/sqrt(x))

To find b, let's evaluate the derivative at x = 1:

3/2 = (b/2)*(1/sqrt(1))
3/2 = b/2

Simplifying, we find b = 3.

Step 4: Find the y-intercept
To find the value of a in the equation f(x) = a + b*sqrt(x), we can use the given point (1, 4).

4 = a + 3*sqrt(1)
4 = a + 3

Simplifying, we find a = 1.

Putting it all together, the function tangent to the line 2y - 3x = 5 at the point (1, 4) is:

f(x) = 1 + 3*sqrt(x)

Now let's move on to the second problem.

For the second problem, we need to find a function of the form f(x) = c + d*cos(kx) that is tangent to the lines y = 1 and y = x + (3/2) - (pi/4) at the given points.

Step 1: Find c
Since the first line is y = 1, we know that the value of c should be 1, as this function is always equal to 1, regardless of x or k.

Step 2: Find d
To find the value of d, we need to find the derivative of the function f(x) = 1 + d*cos(kx) at the point (0, 1), which lies on the line y = 1.

Taking the derivative, we get:

f'(x) = -d*k*sin(kx)

Evaluating f'(x) at x = 0:

f'(0) = -d*k*sin(0)
f'(0) = 0

Since the slope of the line y = 1 is 0 at x = 0, the derivative of the function f(x) = 1 + d*cos(kx) should also be 0 at x = 0. Therefore, we can set f'(0) = 0:

0 = -d*k*sin(0)
0 = -d*k*0
0 = 0

This equation holds true for any value of d and k, so we cannot determine the exact values of d and k from this equation.

Step 3: Find the value of k
To find the value of k, we need to use the fact that the function is also tangent to the line y = x + (3/2) - (pi/4) at ([pi/4], [3/2]).

Similarly, we take the derivative of f(x) = 1 + d*cos(kx) and evaluate it at x = [pi/4]:

f'([pi/4]) = -d*k*sin(k*[pi/4])

Since this derivative should be equal to the slope of the line y = x + (3/2) - (pi/4) at x = [pi/4], we set them equal to each other:

1 = 1 + d*k*sin(k*[pi/4])

Simplifying and canceling out the common terms, we get:

0 = d*k*sin(k*[pi/4])

To solve this equation, we need to find the values of k that make sin(k*[pi/4]) = 0.

Since sin(0) = 0 and sin(n*pi) = 0 for any integer n, we know that k*[pi/4] should be equal to n*pi, where n is an integer.

Setting k*[pi/4] = n*pi, we find that k = (4n)/[pi].

So, the value of k is dependent on the integer n.

In summary, the function of the form f(x) = c + d*cos(kx) that is tangent to the lines y = 1 and y = x + (3/2) - (pi/4) at the given points is:

f(x) = 1 + d*cos((4n*x)/[pi]), where n is any integer.

To find a function that is tangent to a given line at a specific point, you need to find the values of the constants in the function that satisfy the conditions of tangency.

1. Function f(x) = a + b sqrt(x) tangent to the line 2y - 3x = 5 at (1,4):

To find the values of a and b, we need to ensure that the function f(x) and the line have the same slope at the point of tangency. The slope of the line 2y - 3x = 5 can be determined by rearranging the equation in slope-intercept form (y = mx + c) where m is the slope:

2y - 3x = 5
2y = 3x + 5
y = (3/2)x + 5/2

So, the slope of the line is m = 3/2.

To find the slope of the tangent line to f(x), we take the derivative of f(x):

f(x) = a + b sqrt(x)
f'(x) = b / (2 sqrt(x))

Now, equate the slopes:

f'(1) = m
b / (2 sqrt(1)) = 3/2
b/2 = 3/2
b = 3

Next, substitute the given point (1,4) into the original equation f(x) and solve for a:

4 = a + 3 sqrt(1)
4 = a + 3
a = 1

Therefore, the function f(x) = 1 + 3 sqrt(x) is tangent to the line 2y - 3x = 5 at the point (1,4).

2. Function f(x) = c + d cos(kx) tangent to y = 1 at (0,1) and y = x + (3/2) - (pi/4) at ([pi/4], [3/2]):

Similar to the previous case, we need to ensure that the function f(x) and the lines have the same slope at the respective points of tangency.

For the first condition, the function f(x) needs to be tangent to y = 1 at (0,1). The slope of y = 1 is constant and equal to 0. To find the derivative of f(x), we use the chain rule:

f'(x) = -dkdxcos(kx)

Since the slope of y = 1 is 0, the derivative of f(x) evaluated at x = 0 should also be 0:

f'(0) = -dkdxcos(0) = 0
-dkdx = 0
dkdx = 0

This means that k can be any value, as long as dkdx = 0.

For the second condition, the function f(x) needs to be tangent to y = x + (3/2) - (pi/4) at ([pi/4], [3/2]). Again, we find the derivative of f(x) using the chain rule:

f'(x) = -dkdxcos(kx)

Now, we equate the slopes:

f'([pi/4]) = 1
-dkdxcos(k[pi/4]) = 1

Solving this equation for dkdx will give us a value for k.

Once you have the value of k, you can put it back into the original equation f(x) = c + d cos(kx) and solve for c and d using the given conditions of tangency.

Please note that without the specific value of k or additional conditions, the values of c and d cannot be determined exactly.