8. If f(x) = sin(7 − 5x), find f′(π), which is the derivative at π:
−0.754
−0.657
0
* 0.657
3.770
9. Given the function:
g(x)={x+b, x<0
{cos(x), x≥0
Find the value of b, if any, that will make the function differentiable at x = 0.
0
1
* 2
No such value exists.
There is not enough information to determine the value.
8.
f'(x) = -5cos(7-5x)
f'(π) = -5cos(7-5π)
= 3.769
or your last choice of 3.770
#9.first, you need
0+b = cos(0)
so b=1
Now f is continuous
But you also need the derivative to be defined at x=0, so
1 = -sin(0)
But it does not. What that means is the left limit and right limit do not agree, so g' has a break at x=0.
If you plot both parts of g(x), you can see that the slope changes abruptly at x=0:
http://www.wolframalpha.com/input/?i=plot+y%3Dx%2B1+and+y%3Dcos%28x%29
8. To find f'(x), the derivative of f(x), we can use the chain rule. The chain rule states that if we have a composite function y = f(g(x)), then the derivative of y with respect to x is given by dy/dx = f'(g(x)) * g'(x).
In this case, f(x) = sin(7 - 5x). The composite function within sin is g(x) = 7 - 5x. Taking the derivative of g(x) with respect to x, we get g'(x) = -5.
Next, we take the derivative of sin(u), where u is the argument of sin. The derivative of sin(u) with respect to u is cos(u), so f'(g(x)) = cos(7 - 5x).
Now we can multiply f'(g(x)) by g'(x) to find f'(x):
f'(x) = -5 * cos(7 - 5x).
To find f'(π), we substitute π for x in f'(x):
f'(π) = -5 * cos(7 - 5π).
But the given options do not include this value. Therefore, it seems like there might be an error in the options provided.
9. To make the function g(x) differentiable at x = 0, the left-hand limit and the right-hand limit of g(x) as x approaches 0 must be equal.
For x < 0, g(x) = x + b. As x approaches 0 from the left side, g(x) approaches b.
For x ≥ 0, g(x) = cos(x). As x approaches 0 from the right side, g(x) approaches cos(0) = 1.
So in order for g(x) to be differentiable at x = 0, the left-hand limit of g(x) as x approaches 0 (which is b) must be equal to the right-hand limit (which is 1).
Therefore, b = 1.
To solve question 8, we can start by finding the derivative of f(x) with respect to x. The derivative of sin(x) is cos(x). To apply the chain rule, we need to multiply by the derivative of the inner function, which is -5 because the derivative of 7-5x with respect to x is -5:
f'(x) = cos(7 - 5x) * (-5)
Now, we can substitute π into the expression for f'(x):
f'(π) = cos(7 - 5π) * (-5)
To find the numerical value, we need to evaluate cos(7 - 5π). Let's do that:
cos(7 - 5π) = cos(7) * cos(5π) + sin(7) * sin(5π)
Since cos(5π) = cos(π), which is -1, and sin(5π) = sin(π), which is 0, we have:
cos(7 - 5π) = cos(7) * (-1) + sin(7) * 0
= -cos(7)
Finally, we can substitute this value back into the expression for f'(π):
f'(π) = -cos(7) * (-5)
= 5cos(7)
By evaluating cos(7) using a calculator or trigonometric table, we find that cos(7) is approximately 0.753.
Therefore, f'(π) ≈ 5 * 0.753 ≈ 3.765
None of the given options match exactly with this value, but the closest option is 3.770.
For question 9, to make the function g(x) differentiable at x = 0, the derivatives of the two pieces of the function must match at x = 0.
To find the derivative of g(x) for x < 0, which is the left piece of the function, we differentiate x + b with respect to x, which is just 1.
To find the derivative of g(x) for x ≥ 0, which is the right piece of the function, we differentiate cos(x) with respect to x, which is -sin(x).
For the function to be differentiable at x = 0, these two derivatives must match:
1 = -sin(0)
1 = 0
Since 1 does not equal 0, we conclude that there is no value of b that will make the function differentiable at x = 0.
Therefore, the correct answer is "No such value exists."