Answer:
The two horiz. tang. lines here are y = -3 and y = 192.
Step-by-step explanation:
Remember that the slope of a tangent line to the graph of a function is given by the derivative of that function. Thus, we find f '(x):
f '(x) = x^2 + 6x - 16. This is the formula for the slope. We set this = to 0 and determine for which x values the tangent line is horizontal:
f '(x) = x^2 + 6x - 16 = 0. Use the quadratic formula to determine the roots here: a = 1; b = 6 and c = -16: the discriminant is b^2-4ac, or 36-4(1)(-16), which has the value 100; thus, the roots are:
-6 plus or minus √100
x = ----------------------------------- = 2 and -8.
2
Evaluating y = x^3/3+3x^2-16x+9 at x = 2 results in y = -3. So one point of tangency is (2, -3). Remembering that the tangent lines in this problem are horizontal, we need only the y-coefficient of (2, -3) to represent this first tangent line: it is y = -3.
Similarly, find the y-coeff. of the other tangent line, which is tangent to the curve at x = -8. The value of x^3/3+3x^2-16x+9 at x = -8 is 192, and so the equation of the 2nd tangent line is y=192 (the slope is zero).
Answer:
x = π/4.
Step-by-step explanation:
3sin(2x) = 2sin(2x) + 1
3sin(2x) - 2sin(2x) = 1
1sin(2x) = 1
sin(2x) = 1
When a variable n = π/2, sin(π/2) = 1 [refer to the unit circle].
2x = π/2
x = π/4.
Hope this helps!
All you would have to do is substitute!
Answer:
First one will be 2
Second one will be 10
Third one will be undefined
Explanation:
So for the first one you are substituting x for -3 so then the equation would look like this: -3^2+9/-3^2, this will equal to 9+9/9 which your answer would be 2 since 18/9 is 2. For the second one it will look like this -1^2+9/-1^2 which would be 1+9/1, this would equal to 10. For the last one it will look like this 0^2+9/0^2 which would just be 0+9/0 9/0 would be undefined.
