To identify the intervals of the function y = 5x^2 - 41x + 42, we need to determine where the function is increasing or decreasing.
To do this, we can take the derivative of the function, which will give us the slope of the tangent line at any given point.
The derivative of y = 5x^2 - 41x + 42 is y' = 10x - 41.
Setting y' equal to zero and solving for x, we get:
10x - 41 = 0
10x = 41
x = 41/10
This gives us the critical point of the function.
To determine the intervals of increasing and decreasing, we can use test values within each interval.
For values of x less than 41/10 (approximately 4.1), if we plug them into y' = 10x - 41, the result will be negative. This means that the function is decreasing in this interval.
For values of x greater than 41/10, if we plug them into y' = 10x - 41, the result will be positive. This means that the function is increasing in this interval.
Therefore, the function y = 5x^2 - 41x + 42 is decreasing when x is less than 41/10, and it is increasing when x is greater than 41/10.
So the intervals of the function are:
(-infinity, 41/10) for decreasing values of y
(41/10, +infinity) for increasing values of y
identify the intervals of y=5x^2-41x+42
3 answers
find the interval(s) of y=5x^2-41x+42
To find the intervals of the function y = 5x^2 - 41x + 42, we need to determine where the function is increasing or decreasing.
First, let's find the critical points of the function by finding the x-values where the derivative of the function is equal to zero or undefined.
Finding the derivative of y = 5x^2 - 41x + 42:
y' = 10x - 41
Setting y' equal to zero:
10x - 41 = 0
10x = 41
x = 4.1
Since the derivative is always defined for every x-value, we only have one critical point at x = 4.1.
Now, we can determine the intervals of increase and decrease by testing points within each interval.
For x values less than 4.1, let's test x = 0:
y' = 10(0) - 41 = -41
Since the derivative is negative, the function is decreasing in this interval.
For x values greater than 4.1, let's test x = 5:
y' = 10(5) - 41 = 9
Since the derivative is positive, the function is increasing in this interval.
Therefore, we can conclude that the function y = 5x^2 - 41x + 42 is decreasing for x < 4.1 and increasing for x > 4.1.
Thus, the interval(s) for y = 5x^2 - 41x + 42 are:
(-infinity, 4.1) for the interval of decreasing values of y
(4.1, +infinity) for the interval of increasing values of y
First, let's find the critical points of the function by finding the x-values where the derivative of the function is equal to zero or undefined.
Finding the derivative of y = 5x^2 - 41x + 42:
y' = 10x - 41
Setting y' equal to zero:
10x - 41 = 0
10x = 41
x = 4.1
Since the derivative is always defined for every x-value, we only have one critical point at x = 4.1.
Now, we can determine the intervals of increase and decrease by testing points within each interval.
For x values less than 4.1, let's test x = 0:
y' = 10(0) - 41 = -41
Since the derivative is negative, the function is decreasing in this interval.
For x values greater than 4.1, let's test x = 5:
y' = 10(5) - 41 = 9
Since the derivative is positive, the function is increasing in this interval.
Therefore, we can conclude that the function y = 5x^2 - 41x + 42 is decreasing for x < 4.1 and increasing for x > 4.1.
Thus, the interval(s) for y = 5x^2 - 41x + 42 are:
(-infinity, 4.1) for the interval of decreasing values of y
(4.1, +infinity) for the interval of increasing values of y