find the equation of the line tangent to the graph y=3e^x at x=1
Well, you know what they say about tangents and graphs. They're like two peas in a pod, or, in this case, like two mathematical friends having a little rendezvous. Let's find the equation of the tangent line, shall we?
First, let's remind ourselves of a little mathematical trick. The slope of the tangent line to a function at a given point is equal to the derivative of that function evaluated at that point. So, we need to find the derivative of y = 3e^x.
To do that, we'll apply the power rule to the function. The power rule states that if y = ax^n, then the derivative of y with respect to x is given by dy/dx = anx^(n-1). In this case, n is equal to 1, so the derivative of y = 3e^x is 3e^x.
Now that we have the derivative, let's find the slope of the tangent line by evaluating the derivative of y with respect to x at x = 1. Plugging in x = 1, we get dy/dx = 3e^1 = 3e.
So, the slope of the tangent line to the graph y = 3e^x at x = 1 is 3e.
To find the equation of the line, we need a point that lies on the tangent line. We know that the point (1, 3e^1) lies on the original graph y = 3e^x.
Using the point-slope form of a linear equation, which is y - y1 = m(x - x1), we can substitute the slope (m = 3e), the point (x1 = 1, y1 = 3e^1), and solve for y:
y - 3e^1 = 3e(x - 1)
And voila! We have the equation of the line tangent to the graph y = 3e^x at x = 1. I hope this little mathematical rendezvous leaves you with a smile!
To find the equation of the line tangent to the graph of y = 3e^x at x = 1, you need to find the slope of the tangent line and a point on the line.
Step 1: Find the derivative of the function y = 3e^x with respect to x.
Recall that the derivative of e^x is e^x. So, the derivative of 3e^x is 3e^x.
Step 2: Substitute x = 1 into the derivative to get the slope of the tangent line.
When x = 1, the derivative is 3e^1 = 3e. Therefore, the slope of the tangent line is 3e.
Step 3: Find a point on the line using the given function at x = 1.
When x = 1, y = 3e^1 = 3e. So, the point (1, 3e) is on the tangent line.
Step 4: Use the point-slope form of the equation of a line.
The equation of a line with slope m and passing through the point (x1, y1) is given by:
y - y1 = m(x - x1)
Substituting the values we found, the equation of the line tangent to the graph of y = 3e^x at x = 1 is:
y - 3e = 3e(x - 1)
This can be simplified to:
y = 3ex - 3e + 3e
y = 3ex
So, the equation of the line tangent to the graph y = 3e^x at x = 1 is y = 3ex.
To find the equation of the line tangent to the graph of y = 3e^x at x = 1, we need to determine the slope of the tangent line at that point and then use the point-slope form of a linear equation to find the equation of the line.
Step 1: Find the derivative of y = 3e^x with respect to x.
The derivative of y = 3e^x is dy/dx = 3e^x.
Step 2: Evaluate the derivative at x = 1 to determine the slope of the tangent line.
Substituting x = 1 into dy/dx = 3e^x, we get dy/dx|_(x=1) = 3e^1 = 3e.
So, the slope of the tangent line at x = 1 is 3e.
Step 3: Use the point-slope form of a linear equation to find the equation of the tangent line.
The point-slope form of a linear equation is given by y - y1 = m(x - x1), where (x1, y1) is a point on the line and m is its slope.
Since the tangent line passes through the point (1, 3e^1) on the graph y = 3e^x, we can substitute x1 = 1, y1 = 3e into the point-slope equation.
Therefore, the equation of the tangent line is y - 3e = 3e(x - 1).
Simplifying this equation, we get:
y - 3e = 3ex - 3e.
Thus, the equation of the line tangent to the graph y = 3e^x at x = 1 is y = 3ex - 2e.
you have a point (1,3e)
and a slope y'(1) = 3e
so, the line is
y-3e = 3e(x-1)