Let f be the function defined by f(x)=2x+11-3e^x
If the equation of the tangent line to the graph of f at the point (0,8) is given by y=mx+b then:
m=?
and b=?
For the standard line equation,
y=mx+b
m is the slope of the line
b is the y intercept (where x=0)
Take the derivative of the equation. Use that to find the slope of the tangent at (0,8). Once you have the slope and one point (0,8) you can solve for the b.
I just need help finding the derivative of f(x). The 3e^x is tripping me out.
if
y = 3e^x
then
dy/dx = 3 e^x
because
d/dx e^u = e^u du/dx
To find the equation of the tangent line to the graph of f at the point (0,8), we can use the concept of derivatives. The slope of the tangent line is equal to the derivative of the function evaluated at the given point.
First, let's find the derivative of f(x). The function f(x) = 2x + 11 - 3e^x can be written as:
f(x) = 2x + 11 - 3 * e^x.
To find the derivative, we differentiate each term separately:
f'(x) = d/dx (2x) + d/dx (11) - d/dx (3e^x).
= 2 - 0 - 3 * d/dx (e^x).
The derivative of e^x with respect to x is simply e^x:
f'(x) = 2 - 3 * e^x.
Now, let's find the slope (m) of the tangent line at the point (0,8). Since the slope is the derivative evaluated at the given point, we substitute x = 0 (since (0,8) is the given point):
m = f'(0) = 2 - 3 * e^0 = 2 - 3 * 1 = 2 - 3 = -1.
Therefore, the slope of the tangent line is m = -1.
To find the y-intercept (b) of the tangent line, we substitute the coordinates of the given point (0,8) and the slope (-1) into the equation y = mx + b:
8 = (-1) * 0 + b,
8 = b.
Therefore, the y-intercept (b) of the tangent line is b = 8.
Overall, the equation of the tangent line to the graph of f at the point (0,8) is y = -x + 8.