From first principles (ie using the tangent slope method), find the slope of the following curves at the given value of x.
f(x)=2x^2−6x at x = 3
f(x)=2x^2-6x
f(x+h)= 2(x+h)^2-6(x+h)
=2x^2+4xh+2h^2-6x-6h
lim h-->0 f(x+h)-f(x)/h
=2x^2+4xh+2h^2-6x-6h - (2x^2-6x)
=4xh+2h^2-6h/h
=h(4x+2h-6)/h
=lim h -->0 = 4x+2h-6
4x+2(0)-6
=4x-6
at x=3
f(3)=4x-6
=4(3)-6
=6
f(3) = 2(9) - 6(3) = 0
f(3+h) = 2(3+h)^2 - 6(3+h)
= 2(9 + 6h + h^2 - 18 - 6h
= 2h^2 + 6h
slope = lim (f(3+h) - f(3) )/h , as h--->0
= lim(2h^2 + 6h - 0)/h
= lim h(2h + 6)/h
= lim 2h + 6 , as h ---> 0
= 6
either way works fine. Plug in the point before or after. The first way gets you to 4x-6, and then you can use that to find the slope at any point, without having to go through all of the steps each time.
To find the slope of the curve f(x) = 2x^2 - 6x at x = 3 using the tangent slope method from first principles, we start by finding the derivative of f(x) with respect to x.
f'(x) = lim h->0 (f(x+h) - f(x))/h
Let's substitute the values into the above expression.
f(x) = 2x^2 - 6x
f(x+h) = 2(x+h)^2 - 6(x+h)
Expanding the expression of f(x+h):
f(x+h) = 2(x^2 + 2xh + h^2) - 6x - 6h
= 2x^2 + 4xh + 2h^2 - 6x - 6h
Now we can find the difference quotient:
(f(x+h) - f(x))/h = (2x^2 + 4xh + 2h^2 - 6x - 6h - (2x^2 - 6x))/h
= (2x^2 + 4xh + 2h^2 - 6x - 6h - 2x^2 + 6x)/h
= (4xh + 2h^2 - 6h)/h
= (h(4x + 2h - 6))/h
Simplifying by canceling out the h's:
(f(x+h) - f(x))/h = 4x + 2h - 6
Now, to find the slope at x = 3, we substitute x = 3 into the expression above:
Slope at x = 3 = 4(3) + 2(0) - 6
= 12 - 6
= 6
Therefore, the slope of the curve f(x) = 2x^2 - 6x at x = 3 is 6.