Draw a cartesian plane, create a graph with the equation x = y^2 - 2
then substitute numbers into the equation so that it is true, to find points on the graph, e.g. substitute y with 1, you get
x = 1^2 - 2
x = 1 - 2 = -1, so when y = 1, x = -1, this point is (-1, 1)
for the next substitute y with 2,
x = 2^2 - 2
x = 4 - 2 = 2, the point is (2, 2)
you might want to try negative values of y
y = -1, x = (-1)^2 - 2
x = -1 the point is (-1,-1)
then plot the points on the graph
Do you have a picture? i can help if you have one
Yes. A is the right graph because the line must travel through two points: (0 ,1)and(-3, 0)
The given function is
f(x) = x - ln(8x), on the interval [1/2, 2].
The derivative of f is
f'(x) = 1 - 1/x
The second derivative is
f''(x) = 1/x²
A local maximum or minimum occurs when f'(x) = 0.
That is,
1 - 1/x = 0 => 1/x = 1 => x =1.
When x = 1, f'' = 1 (positive).
Therefore f(x) is minimum when x=1.
The minimum value is
f(1) = 1 - ln(8) = -1.079
The maximum value of f occurs either at x = 1/2 or at x = 2.
f(1/2) = 1/2 - ln(4) = -0.886
f(2) = 2 - ln(16) = -0.773
The maximum value of f is
f(2) = 2 - ln(16) = -0.773
A graph of f(x) confirms the results.
Answer:
Minimum value = 1 - ln(8)
Maximum value = 2 - ln(16)