Find all points on the curve here the slope is 1
y = x^4 - 31x + 11
To find all the points on the curve where the slope is 1, we need to find the values of x for which the derivative of y with respect to x is equal to 1.
First, let's find the derivative of y with respect to x:
y = x^4 - 31x + 11
To find the derivative, we differentiate each term one at a time using the power rule:
dy/dx = 4x^3 - 31
Now, let's set this derivative equal to 1 and solve for x:
4x^3 - 31 = 1
Rearranging the equation:
4x^3 = 32
Dividing both sides by 4:
x^3 = 8
Taking the cube root of both sides:
x = 2
So, the value of x where the slope is 1 is x = 2. Now, let's find the corresponding y-value by substituting x = 2 into the original equation:
y = (2)^4 - 31(2) + 11
y = 16 - 62 + 11
y = -35
Therefore, the point on the curve where the slope is 1 is (2, -35).