let f(x)= (g(x)/h(x) where g(2)=18, h(2)=6, g'(2)=5 and h'(2)=2. Find the equation of the normal to the graph of f at x=2.
The slope of the tangent to f(x) is
f'(x) = (g'h-gh')/h^2
So, the normal has slope h^2/(gh'-g'h)
At x=2, that is 6^2/(18*2-5*6) = 6
f(2) = g(2)/h(2) = 18/6 = 3
So, now you have a point and a slope, so the normal line at x=2 is
y-3 = 6(x-2)
Well, well, well, it looks like we have a calculus problem on our hands! Time to put on our mathematical clown shoes and dive in.
To find the equation of the normal to the graph of f at x = 2, we first need to find the slope of the tangent line. The slope of the tangent line is simply the derivative of f evaluated at x = 2.
Now, let's start by finding the derivatives of g(x) and h(x). Since we are given g'(2) and h'(2), we can deduce the values of g(x) and h(x) at x = 2.
Given g(2) = 18, g'(2) = 5, h(2) = 6, and h'(2) = 2, we can find f'(x) using the quotient rule. The quotient rule states that if f(x) = g(x)/h(x), then f'(x) = (g'(x) * h(x) - g(x) * h'(x)) / (h(x))^2.
Now, let's substitute the given values into the quotient rule formula:
f'(x) = (5 * 6 - 18 * 2) / (6)^2
= (30 - 36) / 36
= -6 / 36
= -1/6
Great! We have calculated f'(x) = -1/6 as the slope of the tangent line to f at x = 2.
To find the slope of the normal line (perpendicular to the tangent line), we take the negative reciprocal of -1/6, which gives us the slope of 6/1 or simply 6.
Now, we know the slope of the normal line and a point on the normal line, namely, the point (2, f(2)).
Since f(x) = g(x)/h(x), we can substitute x = 2 into the given function to find f(2):
f(2) = g(2)/h(2)
= 18/6
= 3
Therefore, at x = 2, the point on the normal line is (2, 3), and the slope of the normal line is 6.
Using the point-slope form of a line, the equation of the normal line to the graph of f at x = 2 is:
y - 3 = 6(x - 2)
And if we simplify it:
y - 3 = 6x - 12
Or in a more funky clownee way:
y = 6x - 9
There we go! The equation of the normal to the graph of f at x = 2 is y = 6x - 9. I hope that brings a smile to your face!
To find the equation of the normal to the graph of f at x=2, we need to find the slope of the tangent line to the graph of f at x=2 and then find the negative reciprocal of that slope to get the slope of the normal line.
Step 1: Find the derivative of f(x):
f'(x) = [g'(x)h(x) - g(x)h'(x)] / (h(x))^2
Step 2: Plug in the given values into f'(x):
f'(2) = [g'(2)h(2) - g(2)h'(2)] / (h(2))^2
= [5 * 6 - 18 * 2] / (6)^2
= [30 - 36] / 36
= -6 / 36
= -1/6
Step 3: The slope of the tangent line (m) is f'(2), so m = -1/6.
Step 4: The slope of the normal line is the negative reciprocal of the slope of the tangent line. Let's call it n:
n = -1/m
= -1/(-1/6)
= 6
Step 5: We have the slope of the normal line and the point (2, f(2)). We can write the equation of the normal line using the point-slope form:
y - y1 = n(x - x1)
Using (x1, y1) = (2, f(2)), the equation becomes:
y - f(2) = 6(x - 2)
We still need the value of f(2) to complete the equation.
Step 6: Calculate f(2):
f(2) = g(2) / h(2)
= 18 / 6
= 3
Step 7: Plug in the value of f(2) into the equation to get the final equation of the normal line:
y - 3 = 6(x - 2)
This is the equation of the normal to the graph of f at x=2.
To find the equation of the normal to the graph of f at x = 2, we need to determine the slope and the point at which the normal line intersects with the graph.
First, let's find the slope of the tangent line. The slope of the tangent line can be found by taking the derivative of f(x) and evaluating it at x = 2.
f(x) = g(x)/h(x)
To find the derivative of f(x), we use the Quotient Rule:
f'(x) = (g'(x)h(x) - g(x)h'(x)) / (h(x))^2
Now substitute the given values:
f'(2) = (g'(2)h(2) - g(2)h'(2)) / (h(2))^2
= (5 * 6 - 18 * 2) / 6^2
= (30 - 36) / 36
= -6 / 36
= -1/6
So, the slope of the tangent line at x = 2 is -1/6.
Now, let's find the point at which the line intersects with the graph. To do this, we substitute x = 2 into the equation of f(x):
f(2) = g(2)/h(2)
= 18/6
= 3
Therefore, the point at which the line intersects the graph is (2, 3).
Now we have the slope (-1/6) and a point (2, 3) on the line, which allows us to find the equation of the normal line using the point-slope form.
Using the point-slope form: y - y1 = m(x - x1), where (x1, y1) is the point and m is the slope, we can substitute the values:
y - 3 = (-1/6)(x - 2)
y - 3 = (-1/6)x + 1/3
y = (-1/6)x + 1/3 + 3
y = (-1/6)x + 1/3 + 9/3
y = (-1/6)x + 10/3
Therefore, the equation of the normal to the graph of f at x = 2 is y = (-1/6)x + 10/3.