Step 1: Find f'(x):
f'(x) = -6x^2 + 6x
Step 2: Evaluate f'(2) to find the slope of the tangent line at x=2:
f'(2) = -6(2)^2 + 6(2) = -24 + 12 = -12
Step 3: Find f(2), so you have a point on y=f(x):
f(2) = -2·(2)^3 + 3·(2)^2 = -16 + 12 = -4
So, you have the point (2,-4) and the slope of -12.
Step 4: Find the equation of your tangent line:
Using point-slope form you'd have: y + 4 = -12 (x - 2)
That is the equation of the tangent line.
If your teacher is picky and wants slope-intercept, solve that for y to get:
y = -12 x + 20
The input of each of these functions is always an angle, and as you learned in the previous sections, these angles can take on any real number value. Therefore the sine and cosine function have the same domain, the set of all real numbers, \begin{align*}R\end{align*}
You get it by cross multiplication
10×9÷3=30 and if you want to check
30 over 9 =3.333
10 over 3 = 3.333
Refer to the diagram shown below.
The shaded area satisfies the two inequalities
y > x - 1
and
y < 1 -x
Answer: A and D
Answer:
300%
Step-by-step explanation:
Your technically just multiplying by 3. Hope this helps