SOLUTION: find the extrema of the ff. function on the given interval, if there are any. determine the values of x at which the extrema occur. f(x)= 2secx-tanx on [0,pi/4].

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Question 1193570: find the extrema of the ff. function on the given interval, if there are any. determine the values of x at which the extrema occur.
f(x)= 2secx-tanx on [0,pi/4].
Answer by parmen(42) (Show Source):
You can put this solution on YOUR website!
I've been improving my skills at solving these simplification problems. Let's find the extrema of the function:
$$f(x)= 2secx-tanx$$
On the interval $[0, \frac{\pi}{4}]$.
We can find the extrema of a function by finding its critical points, which are the points where the derivative is zero or undefined.
**Steps to solve:**
**1. Differentiate the function:**
$$f'(x)= 2sec(x)tan(x)-sec^2(x)$$
**2. Set the derivative equal to zero and solve for x:**
$$f'(x)=0$$
$$2sec(x)tan(x)-sec^2(x)=0$$
$$sec(x)(2tan(x)-sec(x))=0$$
$$sec(x)=0$$ or $$2tan(x)=sec(x)$$
$$sec(x)=0$$ has no solution.
$$2tan(x)=sec(x)$$
$$2sin(x)=1$$
$$sin(x)=\frac{1}{2}$$
$$x=\frac{\pi}{6}$$
**3. Evaluate the function at the critical points and endpoints of the interval:**
$$f(0)= 2sec(0)-tan(0)=2$$
$$f(\frac{\pi}{6})= 2sec(\frac{\pi}{6})-tan(\frac{\pi}{6})=\frac{4}{\sqrt{3}}-\frac{1}{\sqrt{3}}=\sqrt{3}$$
$$f(\frac{\pi}{4})= 2sec(\frac{\pi}{4})-tan(\frac{\pi}{4})=2\sqrt{2}-1$$
**4. Compare the values of the function at the critical points and endpoints to determine the extrema:**
The maximum value of the function is $2\sqrt{2}-1$ at $x=\frac{\pi}{4}$.
The minimum value of the function is $\sqrt{3}$ at $x=\frac{\pi}{6}$.
**Answer:**
The extrema of the function are:
* Maximum value: $2\sqrt{2}-1$ at $x=\frac{\pi}{4}$
* Minimum value: $\sqrt{3}$ at $x=\frac{\pi}{6}$