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If f'(x)=sin^3 (x) - cos^4 (x) , determine where the graph of f(x) is increasing on the interval [0,10].

This is one of my AP calculus summer homework questions and I have no idea where to begin... Any help you could give me would be greatly appreciated! Thank you!

To determine where the function increasing / decreasing, the derivative needs to be looked at.

If f(x) = f'(x) = sin^3(x) - cos^4(x), take f(x) = m^3(x) - n^4(x).

From this, we can say that f'(x) = 3m^(x)m'(x) - 4n^3(x)n'(x).

Since m(x) = sin(x), m'(x) = cos(x); since n(x) = cos(x), n'(x) = -sin(x).

This gives f'(x) = 3*sin^2(x)cos(x) + 4*cos^3(x)sin(x).

Factoring out sin(x)cos(x), we get f'(x) = sin(x)cos(x)[3sin(x) + 4cos^2(x)].

Looking at each part, we need to know when sin(x) = 0, cos(x) = 0, and

3*sin(x) + 4*cos^2(x) = 0.

It is known that sin(x) = 0 at x = nπ, n = 0, ±1π, ±2π, ...

It is known that cos(x) = 0 at x = (n + 0.5)π, n = 0, ±1π, ±2π, ...

It is known that 3sin(x) + 4cos^2(x) = 0 where 3sin(x) = - 4cos^2(x).

Dividing both sides by 3sin(x) gives 1 = -(4/3)cos^2(x)/sin(x).

Remembering cos(x)/sin(x)= cot(x), this is 0 = (-4/3)cos(x)cot(x),

which is 0 when cos(x) = 0 and cot(x) = 0. Note that cot(x) is undefined where sin(x) = 0,

so the x's are already in the list. Also note that cot(x) = 0 where cos(x) = 0, and we already have those.

Thus, the slope is to be determined in the intervals between x = 0.5nπ, n = 0, ±1π, ±2π, ...

At the π/2 values between the π values, the function is still sloping the same way.

Looking at nπ, the function changes slope at these values.

Looking at the graph in Excel of the function, it starts out at -1 with x=0.

At x = π, the function slopes upward. It levels off there and goes back down to f(x) = -1

at x = 2π.

Calculus

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