Answer:
B. 3.
Step-by-step explanation:
OK lets try again.
The slope of the secant = slope of the tangent at a certain point ( The Mean Value Theorem).
Slope of the secant = f(5) - f(2) / (5 - 2)
= [(25-3) / (5-1) - (4-3) / (2-1)] / 3
= (22/4 - 1) / 3
= 9/2 / 3
= 9/6
= 3/2.
The derivative at c = the slope of the tangent at c.
Finding the derivative:
f'(x) = [2x(x - 1) - (x^2 - 3) ]/ (x - 1)^2 (where x = c).
= (x^2 - 2x + 3)/ (x - 1)^2 = the slope.
So equating the slopes:
(x^2 - 2x + 3) / (x - 1)^2 = 3/2
2x^2 - 4x + 6 = 3x^2 - 6x + 3
x^2 - 2x - 3 = 0
(x - 3)(x + 1) = 90
x = 3 , -1
x can't be -1 because we are working between the values 2 and 5 so
x = c = 3.
Answer:
Todd.
Step-by-step explanation:
The equation has no solutions because negative 2 ≠ 4
Answer: y=-2/5x+2
Solving Steps:
2x+5y=10 - Move the variable to the right
5y=10-2x - Divide both sides by five
y=2-2/5x - Reorder the terms
y=-2/5x+2
The minimum value for g(x)=x² - 10x + 16 is -9
<h3>How to determine the minimum value?</h3>
The function is given as:
g(x)=x² - 10x + 16
Differentiate the function
g'(x) = 2x - 10
Set the function 0
2x - 10 = 0
Add 10 to both sides
2x = 10
Divide by 2
x = 5
Substitute 5 for x in g(x)
g(5)=5² - 10*5 + 16
Evaluate
g(5) = -9
Hence, the minimum value for g(x)=x² - 10x + 16 is -9
Read more about quadratic functions at:
brainly.com/question/7784687
