Calculus Trig EVP

Question

Find the absolute minimum value of

f(x) = 4sin^2(x) + 5cos^2(x)

on [-pi, pi]

Answer 1

Use elementary methods whenever convenient. From the Pythagorean identity sin²x + cos²x = 1, your function becomes f(x) = 4 + cos²x. If cosx takes the value zero at any x in the interval [-π,π], which it does when x is -π/2 or π/2, then clearly we get a minimum value of 4.

Answer 2

The bottom line is: Take the derivative and set it equal to zero to find local minimums and then check the endpoints and find the one that has the lowest value.

For f(x) = 4sin²(x) + 5cos²(x) the derivative is perhaps easier to find if we write it like this:

f(x) = 4(sin(x))² + 5(cos(x))²

To find the derivative, we need to use the power rule AND the chain rule so:

f '(x) = 8sin(x)cos(x) - 10sin(x)cos(x) = -2sin(x)cos(x)

Setting this equal to zero:

-2sin(x)cos(x) = 0 would occur if EITHER sin(x) = 0 or cos(x) = 0

sin(x) = 0 (on the interval [-π, π] occurs at x = -π, x = 0, and x = π

cos(x) = 0 (on [-π, π] occurs at x = -π/2 and x = π/2

So I can build a little table with these x values and plug them in to see wat f(x) is for each of them:

x f(x)

-------------

-π 5

-π/2 4

0 5

π/2 4

π 5

Notice I didn't need to separately add in the endpoints because they were critical points.

So the absolute minimum occurs at 2 points: x = -π/2 and x = π/2 and the value at both is 4

The answer to the question is 4 (it is the absolute minimum value)