if f is a continuous function defined by f(x)={x^2+bx, 5 sin("pie"/2x) then b= ?

f(x)= x^2+bx (for all values of x less than or equal to 5)

f(x)= 5 sin ((pie/2)x) (for all values of x greater than 5)
-------------------------
If it is continuous then those two functions are the same at x = 5
5^2 + 5b = 5 sin (2.5 pi)
5 + b = sin(2.5 pi)
5 + b = sin (450 degrees) = sin(90) = 1
b = 1-5 = -4

I do not understand the { symbol at the beginning and why you have two functions of x, separated by a comma.

Are you asking where the two functions intersect?

Moreover do you really mean

sin("pie"/2x)
or do you mean
sin("pie"x/2)
?

thank you!!

Well, let me get my clown wig on for this one!

To find the value of b, we can start by looking at the two pieces of the function. The first piece, f(x) = x^2 + bx, is a quadratic function. The second piece, f(x) = 5 sin(π/2x), is a sinusoidal function.

Since f(x) is defined to be continuous, we need the two pieces to "connect" smoothly at the points where they meet. In other words, the limit of f(x) as x approaches 0 must be the same from both sides.

For the quadratic function, as x approaches 0, we have f(x) = (0)^2 + b(0) = 0.

For the sinusoidal function, as x approaches 0, we have f(x) = 5 sin(π/2(0)) = 0.

So, in order for these two pieces to connect smoothly at x = 0, we need b to be 0.

Therefore, the value of b is 0.

To find the value of b in the function f(x) = x^2 + bx or f(x) = 5 sin(π/2x), we need to use the definition of continuity of a function.

For a function to be continuous at a point "a", it must satisfy three conditions:

1. The function must be defined at "a".
2. The limit of the function as it approaches "a" from the left side must exist and be equal to the value of the function at "a".
3. The limit of the function as it approaches "a" from the right side must exist and be equal to the value of the function at "a".

First, let's consider the function f(x) = x^2 + bx. To find the value of b, we can evaluate the limit of f(x) as x approaches 0.

lim x→0 (x^2 + bx) = lim x→0 (x^2) + lim x→0 (bx)

The first term evaluates to 0 because any number squared and multiplied by 0 is 0. So, we are left with:

0 + lim x→0 (bx)

To satisfy the continuity condition, the limit of bx must also be 0. This means that for any value of x, the function bx should approach 0 as x approaches 0.

Since bx approaches 0 as x approaches 0 for any value of b, there is no specific value of b that satisfies the condition. Therefore, b can be any real number.

Now, let's consider the function f(x) = 5 sin(π/2x). In this case, we need to evaluate the limit as x approaches 0 again.

lim x→0 (5 sin(π/2x))

To satisfy the continuity condition, the limit of 5 sin(π/2x) must exist as x approaches 0.

However, when x approaches 0, the argument of sin(π/2x) becomes undefined (because sin(π/2*0) is undefined), and therefore, the limit does not exist.

Since the limit of the function does not exist, there is no value of b that satisfies the continuity condition for the function f(x) = 5 sin(π/2x).

In conclusion, for the function f(x) = x^2 + bx, b can be any real number. For the function f(x) = 5 sin(π/2x), no value of b satisfies the continuity condition.

re written:

f(x)= x^2+bx (for all values of x less than or equal to 5)
f(x)= 5 sin ((pie/2)x) (for all values of x greater than 5)

find the value of b