What is the extreme y=4sin(2pi(x+(-45)))+(-2) ?

The extreme y=4sin(2pi(x+(-45)))+(-2) is Maximum(1/4+1*n,2),Minimum(3/4+1*n,-6)

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extreme y=4sin(2pi(x+(-45)))+(-2)

extreme points

y = 4 sin (2 π (x + (- 45))) + (- 2)

Maximum (\frac{1}{4} + 1 \cdot n, 2), Minimum (\frac{3}{4} + 1 \cdot n, - 6)

Solution steps

Find the critical points:

x = \frac{1}{4} + 1 \cdot n, x = \frac{3}{4} + 1 \cdot n

Domain of

4 sin (2 π (x - 45)) - 2 : - \infty < x < \infty

Find intervals:

Increasing

: 1 \cdot n \leq x < \frac{1}{4} + 1 \cdot n,

Decreasing

: \frac{1}{4} + 1 \cdot n < x < \frac{3}{4} + 1 \cdot n,

Increasing

: \frac{3}{4} + 1 \cdot n < x < 1 + 1 \cdot n

Plug the extreme point

x = \frac{1}{4} + 1 n

into

4 sin (2 π (x - 45)) - 2 \Rightarrow y = 2

Maximum (\frac{1}{4} + 1 \cdot n, 2)

Plug the extreme point

x = \frac{3}{4} + 1 n

into

4 sin (2 π (x - 45)) - 2 \Rightarrow y = - 6

Minimum (\frac{3}{4} + 1 \cdot n, - 6)

Maximum (\frac{1}{4} + 1 \cdot n, 2), Minimum (\frac{3}{4} + 1 \cdot n, - 6)

f (x) = x^{4} + 6 x^{5} - x

f (x, y) = \frac{10}{3} x y + x y^{2}

f (x) = - 2 x^{2} + 8 x - 1

f (x) = 4 cos (2 x) - 4

h (t, v) = - 16 t^{2} + v t + c

What is the extreme y=4sin(2pi(x+(-45)))+(-2) ?
The extreme y=4sin(2pi(x+(-45)))+(-2) is Maximum(1/4+1*n,2),Minimum(3/4+1*n,-6)