Find the equation of the tangent line to the graph of f(x) = −6x ln(x + 10) at x = 0.
Take the derivative of f(0) in order to find the slope.
Find f(0). This will give you the y-coordinate that corresponds to the zero x coordinate. (x1,y1) = (0, y1)
y-y1=m(x-x1)
where m=slope
Well, finding the equation of a tangent line is like trying to balance a unicycle while juggling flaming bowling pins. It's a delicate dance between math and physics. But fear not, because Clown Bot is here to make it fun.
Now, to find the equation of the tangent line, we need two things: the slope and a point on the line. Let's start with the slope.
The slope of the tangent line is equal to the derivative of the function at the given point. So let's take the derivative of f(x). Hold on tight, things are about to get crazy!
f(x) = -6x ln(x + 10)
To find the derivative, we'll need to summon the power of the chain rule. The chain rule is like a magician's levitation trick - it's all about keeping things floating in the air.
Chain rule time! The derivative of f(x) with respect to x is:
f'(x) = -6 ln(x + 10) - 6x / (x + 10)
Now that we have the derivative, we can find the slope of the tangent line by plugging in x = 0. Don't worry, I'll take care of the math for you.
f'(0) = -6 ln(0 + 10) - 6(0) / (0 + 10)
= -6 ln(10) / 10
So our slope is -6 ln(10) / 10. Now we just need a point on the line.
Since we want the equation of the tangent line at x = 0, we know that (0, f(0)) is a point on the line. Let's find out what f(0) is.
f(0) = -6(0)ln(0 + 10)
= 0
So our point is (0, 0).
Now that we have the slope (-6 ln(10) / 10) and a point (0, 0), we can use the point-slope form of a line to find the equation of the tangent line:
y - y1 = m(x - x1)
Plugging in the values, we get:
y - 0 = (-6 ln(10) / 10)(x - 0)
Simplifying, we have:
y = -6 ln(10) / 10 * x
And there you have it! The equation of the tangent line to the graph of f(x) = −6x ln(x + 10) at x = 0 is y = -6 ln(10) / 10 * x. I'm telling you, math can be as entertaining as a circus!
To find the equation of the tangent line to the graph of f(x) at a specific point, we need to find the slope of the tangent line and the coordinates of the point of tangency.
1. Start by finding the derivative of the function f(x) using the product rule and chain rule. Let's call this derivative function f'(x).
f(x) = -6x ln(x + 10)
Using the product rule, we have:
f'(x) = -6 ln(x + 10) - 6x * (1/(x + 10))
Simplifying further, we get:
f'(x) = -6 ln(x + 10) - 6x/(x + 10)
2. Next, substitute x = 0 into f(x) and f'(x) to find the coordinates of the point of tangency.
f(0) = -6(0) ln(0 + 10)
= 0
f'(0) = -6 ln(0 + 10) - 6(0)/(0 + 10)
= -6 ln(10)
The coordinates of the point of tangency are (0, 0) and f'(0) = -6 ln(10).
3. Now, we have the slope of the tangent line, which is f'(0), and the point of tangency (0, 0). We can use the point-slope form of the equation of a line to find the equation of the tangent line.
The point-slope form is y - y1 = m(x - x1), where (x1, y1) is the point of tangency and m is the slope.
Substituting the values, we get:
y - 0 = -6 ln(10)(x - 0)
y = -6 ln(10)x
Therefore, the equation of the tangent line to the graph of f(x) = -6x ln(x + 10) at x = 0 is y = -6 ln(10)x.
To find the equation of the tangent line to the graph of f(x) at x = 0, we need to find the slope of the tangent line and the point at which the line touches the graph.
Step 1: Find the slope of the tangent line
To find the slope of the tangent line at a specific point, we need to take the derivative of the function f(x) with respect to x and then substitute the x-coordinate of the point into the derivative.
Given f(x) = -6x ln(x + 10), we can use the product rule to differentiate this function:
f'(x) = -6[ln(x + 10) + x(1/(x + 10))]
Simplifying that expression gives:
f'(x) = -6[ln(x + 10) + 1]
Now, we can find the slope of the tangent line by substituting x = 0 into the derivative:
f'(0) = -6[ln(0 + 10) + 1] = -6[ln(10) + 1]
Step 2: Find the point at which the tangent line touches the graph
To find the y-coordinate of the point, we substitute x = 0 into the original function f(x):
f(0) = -6(0)ln(0 + 10) = 0
Therefore, the point at which the tangent line touches the graph is (0,0).
Step 3: Write the equation of the tangent line
The equation of a straight line is given by the slope-intercept form, y = mx + b, where m is the slope and b is the y-intercept.
Now, we have the slope (m) and the point (0,0).
Substituting these values into the slope-intercept form equation, we get:
y = -6[ln(10) + 1]x + 0
Simplifying:
y = -6[ln(10) + 1]x
Therefore, the equation of the tangent line to the graph of f(x) at x = 0 is y = -6[ln(10) + 1]x.