for function f(x)=4x-2x^3
a.) Find the general formula for the slope of the tangent line using the definition of derivative.
b.) Find the slope at x=1
So for part (a), do you know how to find the derivative of f(x) = 4x-2x^3
yea I actually just figured out part a
now I just need help with b
I got f'(x)=-6x^2+4
do I just plug 1 in for x into -6x+4
You have already gotten your slope
Which is give as Δy/Δx
Just plug 1 into it and you're done
a.) To find the general formula for the slope of the tangent line using the definition of the derivative, we need to apply the definition of derivative, which is the limit of the difference quotient as the change in x approaches zero.
The definition of the derivative of a function f(x) is given by:
f'(x) = lim(h->0) [f(x+h) - f(x)] / h
Applying this definition to the function f(x) = 4x - 2x^3, we can find the general formula for the slope of the tangent line. Let's start by applying the difference quotient:
f'(x) = lim(h->0) [(4(x+h) - 2(x+h)^3) - (4x - 2x^3)] / h
Expanding and simplifying:
f'(x) = lim(h->0) [4x + 4h - 2(x^3 + 3x^2h + 3xh^2 + h^3) - 4x + 2x^3] / h
f'(x) = lim(h->0) [4h - 6x^2h - 6xh^2 - 2h^3] / h
Canceling out the h in each term:
f'(x) = lim(h->0) 4 - 6x^2 - 6xh - 2h^2
Taking the limit as h approaches zero:
f'(x) = 4 - 6x^2
Hence, the general formula for the slope of the tangent line for the function f(x) = 4x - 2x^3 is f'(x) = 4 - 6x^2.
b.) To find the slope at x = 1, we can substitute x = 1 into the general formula for the slope of the tangent line, which we found in part (a):
f'(1) = 4 - 6(1)^2
f'(1) = 4 - 6
f'(1) = -2
Therefore, the slope of the tangent line at x = 1 is -2.