From first principles (ie using the tangent slope method), find the slope of the following curves at the given value of x.
y=3x^3+1 at x=-4
he means
definition of derivative (slope at point x) is
[f(x+h) - f(x) ] / h
as h approaches zero
What do you mean by dividing with h?
To find the slope of the curve using the tangent slope method from first principles, we will differentiate the function with respect to x.
First, let's find the derivative of the function y = 3x^3 + 1 using the power rule for differentiation. The power rule states that if we have a function of the form f(x) = ax^n, then its derivative is given by f'(x) = nax^(n-1).
In this case, the function is y = 3x^3 + 1. Applying the power rule, we get:
y' = 3 * 3x^(3-1) = 9x^2
Now that we have the derivative of the function, we can find the slope at a particular x-value by substituting that value into the derivative.
Given x = -4, we can calculate the slope by substituting -4 into the derivative:
y' = 9(-4)^2 = 9 * 16 = 144
Therefore, the slope of the curve y = 3x^3 + 1 at x = -4 is 144.
f(x+h) - f(x) = (3(x+h)^3+1) - (3x^3-1) = 9x^2h +9xh^2 + 3h^3
Now divide that by h and take the limit as h→0 and you get
y' = 9x^2
now plug in x = -4