2. Make a number line with values greater than and less than the x value you found. Plug and chug and find whether it is positive or negative. If it goes negative to positive, it is increasing. Positive to negative is decreasing.
3. Take the derivative of the outside function multiplied by the inside function multiplied by the derivative of inside function.
f(x)=(4-2x)^3
f"=3(4-2x)^2*-2 = -6(4-2x)^2
Tangent line at x=3
m=-24 (Plug 3 into derivative)
y= -8 (Plug 3 into original equation)
y+8=-24(x-3)
4. h'(-4)=f'(g(-4))g'(-4)
h'(-4)=f'(-5)g'(-4)
h'(-4)=20*2=40
For number 1, you could mention that the points could be maximums or minimums.
ReplyDeleteI agree with Swanson because a critical point is a potential minimum or maximum.
ReplyDeleteI agree with Swanson because a critical point is a potential minimum or maximum.
ReplyDeleteWhere are you plugging and chugging in the second part to find where it is increasing/decreasing????
ReplyDelete