Finding the max/min of expressions such as sin^2(x)+4, cos^2(x)+2cos(x)+6?

Question

My book has horrible explanations and there are only two, i assume for questions such as the second one in my title i would have to simplify the equation, solve it and then find the max/min, but how about for expressions such as sin^2(x)+4, cos^2(x)...etc.

If someone can hint me or teach me a general algebraic rule or such to find max/min's of such equations i would be grateful.

thank you.

Answer 1

sin^2(x)+4 = (1-cos2x)/2 = 4 = 9/2 - cos2x/2

max = 5

min=4

Answer 2

sin^2(x) + 4

should have a max when sin ² x is max or (sin x) has a max or sin x = 1

hence max value sin^2(x) + 4 = 1² + 4 = 5 ---------ans



sin^2(x) + 4 = 1 - cos²x + 4 = 5 - cos ² x

shouls ahev min as max of cos²x or cosx = 1

value is 5 - cos ² x = 5 - 1 = 4 --------ans

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cos^2(x)+2cos(x)+ 6 = (cos x + 1)² + 5 has max at cosx = 1

or value is [1 + 1]² + 5 = 9 ---------ans

(cos x + 1)² + 5 = [2 cos² (x/2)]² + 5 = 2² [ 1 - sin² x/2)² + 5

at sin²(x/2) = +1 this function will attain a minimum

so minimum is 2² [ 1 -1} + 5 = 5 ------------ ans







Ofcourse derivative = 0 also works for both if we find d²/dx² and fidn max or min

Answer 3

To find the maximum and minimum of an expression, set the derivative of the expression equal to zero.



let y = sin²(x) + 4

y' = 2sin(x)cos(x) = 0

y" = 2[cos²(x) - sin²(x)]

sin(x) = 0 => x = 0, 180

cos(x) = 0 => x = 90, 270



The expression evaluated at 0, 180, 90, and 270 will be the maximum and minimum values.

the second derivative will be negative at the maximum values, and positive at the minimum values

Since y" = 2[cos²(x) - sin²(x)] is positive at 0 and 180 and negative at 90 and 270, the values at 0 and 180 are minimums and the values at 90 and 270 are maximums.



let y = cos^2(x)+2cos(x)+6

y' = -2cos(x)sin(x) - 2sin(x) = 0

y" = -2(cos²(x) - sin²(x) - 2cos(x)

y" = 2[sin²(x) - cos²(x) + cos(x)]



sin(x)[cos(x) + 1] = 0

sin(x) = 0 => x = 0, 180

cox(x) = -1 => x = 180

The maximum and minimum occur at 0 and 180.

Answer 4

the general rule for finding the max/min for functions is to calculate the first derivative



The min/max points of the function are those where the first derivative is 0.



To find out if the point is min of max (since the first derivative is 0 for both) you look at the values of the derivative before and after the 0 value.



Let's take the next example:



f . . . . . . . . . . . . . . x1 . . . . . . . . . . . . . . . x2

f' . . (neg. values) . . 0 . . . . . (pos. values) . . 0 . . . (neg. values)



a) if the first derivative has negative values before the investigated value (min/max) and positive values after (see x1) then the point is a min one (the function is first descending - negative values for the first derivative, and then ascending - positive values for the first derivative) then the point will be a min value

b) if the values before are positive and after are negative then the point is a max value (see x2)