How would you solve the following problems?

Question

Ok... i'm completely lost on this and if anyone could help show me how to do the following problems and what the answers would be, it would be so greatly appreciated!!! Thank you!

1)Find an equation of the tangent line to the graph of y = g(x) at x = 3 if g(3) = -6 and g'(3) = 4.

2) a. If f(x) = 4x^2 - 2x, find f '(1)
b. Use f '(1) to find an equation of the line tangent to the curve y= 4x^2 - 2x at the point (1,2)

3) If f(t) = t^3 - 8t, find f '(a)

Answer 1

1) since you have g(3) = -6, this indicates that you have point (3,-6) and also you have g'(3) = 4, and this let you know that the gradient of the line at x - 3 is 4. Therefore we can find the equation of the tangent line of graph y = g(x) at x = 3

using (y-y1) = m(x-x1)
y-(-6) = 4(x-3)
y+6 = 4x -12
y = 4x-18

2)differentiate f(x) and you will get f'(x) = 8x -2
substitute x = 1, and you will get 6, indicate that the gradient is 6 at x = 1.

b) similiar to question 1, you have point (1,2) and gradient at point x = 1, use the same method to find the equation of tangent line.

3) differentiate f(t) and you will get f'(t) = 3t^2 -8
substitute t = a and you will get 3a^2-8

Answer 2

The key here is to realize that g'(3) is the slope of the line at x=3.

so, considering the tangent line in form y=mx+b,

y=4x+b

since g(3)=-6, we have the ordered pair (3,-6) which is on both the tangent line and the function g(x) at x=3 so,

(-6)=4(3)+b
-6=12+b
-18=b

so the whole equation is y=4x-18


hope that helps
be sure and check my arithmetic

Answer 3

f(x)=4x^2-2x
f'(x)=8x-2
f'(1)=8(1)-2
f'(1)=6

I don't remember how to find tangent lines at this moment in time...but I'll think about it