The value of the derivative at the maximum or minimum for a continuous function must be zero.
<h3>What happens with the derivative at the maximum of minimum?</h3>
So, remember that the derivative at a given value gives the slope of a tangent line to the curve at that point.
Now, also remember that maximums or minimums are points where the behavior of the curve changes (it stops going up and starts going down or things like that).
If you draw the tangent line to these points, you will see that you end with horizontal lines. And the slope of a horizontal line is zero.
So we conclude that the value of the derivative at the maximum or minimum for a continuous function must be zero.
If you want to learn more about maximums and minimums, you can read:
brainly.com/question/24701109
If tan theta is -1, we know immediately that theta is in either Quadrant II or Q IV. We need to focus on Q IV due to the restrictions on theta.
Because tan theta is -1, the ray representing theta makes a 45 degree angle with the horiz axis, and a 45 degree angle with the negative vert. axis. Thus the hypotenuse, by the Pythagorean Theorem, tells us that the hyp is sqrt(2).
Thus, the cosine of theta is adj / hyp, or +1 / sqrt(2), or [sqrt(2)]/2
The secant of theta is the reciprocal of that, and thus is
2 sqrt(2)
---------- * ------------ = sqrt(2) (answer)
sqrt(2) sqrt(2)
Answer:
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Step-by-step explanation:
