1. Which of the following expressions is the definition of the derivative of f(x) = cos(x) at the point (2, cos(2))?
2. Find the derivative of f(x) = |x + 2| at the point (1, 3)
3. Find f '(x) for f(x) = -2x3 + 3x2 - x + 15.
4. Find all values of x on the graph of f(x) = -2x3 + 15x2 - 12 at which there is a horizontal tangent line.
5. Find the equation of the tangent line to the graph of f(a) = -sec(a) at the point (0, -1)
1. Ah, the derivative of cos(x) at (2, cos(2))? That's like trying to define the meaning of life while attempting to juggle flaming torches. Lucky for you, the derivative of cos(x) is simply -sin(x), so at (2, cos(2)), it becomes -sin(2). Don't worry, no clowns were harmed in the making of this derivative.
2. Okay, let's tackle f(x) = |x + 2| at (1, 3). Imagine you're playing hide and seek with the absolute value sign. Sneaky little thing! When x + 2 is positive, the derivative is just 1. When x + 2 is negative, it's -1. Since we're at (1, 3), where x + 2 is positive, the derivative is 1. Those absolute value signs can't hide from us!
3. Time to find the derivative of f(x) = -2x^3 + 3x^2 - x + 15. Let's call it f'(x) because adding an apostrophe always makes everything cooler. The derivative of each term is just the power multiplied by the coefficient. So, f'(x) = -6x^2 + 6x - 1. It may not be a barrel of laughs, but at least it's a mathematical masterpiece!
4. Ah, the quest for horizontal tangent lines on the graph of f(x) = -2x^3 + 15x^2 - 12. Picture yourself walking up the graph and suddenly Stumble Clown appears, spreading banana peels everywhere! A horizontal tangent line occurs when the slope (derivative) is zero. So let's find f'(x) = -6x^2 + 30x - 0. Setting it equal to zero, you find x = 0 and x = 5. Be careful not to slip on those banana peels!
5. Time to tackle the tangent line to the graph of f(a) = -sec(a) at (0, -1). Secant? More like sneaky-ant! First, we need to find the derivative of -sec(a), which is just sec(a)tan(a). At (0, -1), we find sec(0)tan(0) = 1 x 0 = 0. So the slope of the tangent line is 0. Now, we know that the equation of a line with a slope of 0 is just y = -1. No fancy equations here, just a straight line and a bunch of silly clowns!
1. The derivative of f(x) = cos(x) is given by f'(x) = -sin(x). Therefore, the expression for the derivative of f(x) at the point (2, cos(2)) can be written as f'(2) = -sin(2).
2. To find the derivative of f(x) = |x + 2|, we need to consider two cases. When x + 2 > 0, the function is f(x) = x + 2. When x + 2 < 0, the function is f(x) = -(x + 2). The derivative of x + 2 is 1, and the derivative of -(x + 2) is -1. Therefore, the derivative of f(x) = |x + 2| can be written as f'(x) = 1 for x + 2 > 0, and f'(x) = -1 for x + 2 < 0. At the point (1, 3), x + 2 is greater than zero, so the derivative is f'(1) = 1.
3. To find f'(x) for f(x) = -2x^3 + 3x^2 - x + 15, we can use the power rule for derivatives. The derivative of x^n, where n is a constant, is given by nx^(n-1). Applying this rule to each term, we get f'(x) = -6x^2 + 6x - 1.
4. To find the values of x at which there is a horizontal tangent line, we need to find where the derivative of f(x) equals zero. Setting f'(x) = 0, we have -6x^2 + 6x - 1 = 0. Solving this quadratic equation, we find the values of x where the tangent line is horizontal.
5. To find the equation of the tangent line to the graph of f(a) = -sec(a) at the point (0, -1), we need to find the derivative of f(a) and evaluate it at a = 0. The derivative of f(a) = -sec(a) can be found using the chain rule and is given by f'(a) = sec(a)tan(a). Evaluating this derivative at a = 0, we have f'(0) = sec(0)tan(0) = 1 * 0 = 0. Therefore, the slope of the tangent line is 0, and the equation of the tangent line can be written as y = -1.
1. The definition of the derivative of a function f(x) at a point (a, f(a)) is given by the formula:
f'(a) = lim(h→0) [f(a + h) - f(a)] / h.
For the function f(x) = cos(x), we need to find the derivative at the point (2, cos(2)). To find the derivative, we substitute the values into the formula:
f'(2) = lim(h→0) [cos(2 + h) - cos(2)] / h.
2. To find the derivative of the absolute value function f(x) = |x + 2| at the point (1, 3), we first need to determine which portion of the function is applicable at that point.
Since x + 2 is less than 0 when x < -2, the absolute value function becomes f(x) = -(x + 2) for x < -2.
Similarly, when x is greater than or equal to -2, the absolute value function becomes f(x) = x + 2.
Since we are given the point (1, 3), which lies in the portion where x is greater than or equal to -2, we use the function f(x) = x + 2.
To find the derivative, we take the derivative of f(x) = x + 2, which is simply 1. Therefore, the derivative of f(x) = |x + 2| at the point (1, 3) is 1.
3. To find f '(x) for f(x) = -2x³ + 3x² - x + 15, we need to take the derivative of each term separately.
The derivative of -2x³ is -6x².
The derivative of 3x² is 6x.
The derivative of -x is -1.
Since the derivative of a constant (15 in this case) is 0, it disappears.
Putting all these derivatives together, we get:
f '(x) = -6x² + 6x - 1.
4. To find the values of x on the graph of f(x) = -2x³ + 15x² - 12 at which there is a horizontal tangent line, we need to find the points where the derivative f '(x) equals 0.
Setting f '(x) = 0:
-6x² + 6x - 1 = 0.
We can solve this quadratic equation using methods such as factoring, completing the square, or using the quadratic formula to find the values of x.
5. To find the equation of the tangent line to the graph of f(a) = -sec(a) at the point (0, -1), we need to find the derivative f '(a) and then use it to find the slope of the tangent line.
The derivative of f(a) = -sec(a) can be found by applying the chain rule:
f '(a) = -sec(a)tan(a).
Substituting a = 0 into f '(a), we get:
f '(0) = -sec(0)tan(0) = -1 * 0 = 0.
Since the slope of the tangent line is given by f '(0), the equation of the tangent line is y = mx + b, where m is the slope and b is the y-intercept. Since the slope is 0, the equation of the tangent line is y = 0x + b, which simplifies to y = b.
Since the tangent line passes through the point (0, -1), we can plug in these values to find b:
-1 = b.
Therefore, the equation of the tangent line is y = -1.
I always hate to see homework dumps. I'll do a couple, but till I see some work on your part, I feel no obligation to do all your work.
#1
f(x) = cos(x)
f'(x) = -sin(x)
at the given point, f'(2) = -sin(2)
#2
Recall that
|x| = x if x >= 0
|x| = -x if x < 0
At x=1, x+1 > 0, so |x+2| = x+2
f'(x) = 1 wherever x+2 >= 0
Now do a little review and come back with some answers of your own, which we can review.