To find the equation of the tangent line at the point (a, f(a)), we will need to find the derivative of the function at x = a.
First, let's find the derivative of f(x):
f(x) = 7e^x + 8x
f'(x) = 7e^x + 8
Next, let's find the derivative at x = a:
f'(a) = 7e^a + 8
Now, we can write the equation of the tangent line using the point-slope form:
y - f(a) = f'(a)(x - a)
Substituting in the values:
y - f(a) = (7e^a + 8)(x - a)
Since a = 0, we have:
y - f(0) = (7e^0 + 8)(x - 0)
Simplifying:
y - f(0) = (7 + 8)(x - 0)
The constant term f(0) can be found by substituting x = 0 into the original function:
f(0) = 7e^0 + 8(0) = 7
So the equation of the tangent line is:
y - 7 = 15x
Find an equation of the line tangent to the following curve at the point (a,f(a)).
f(x)=7e^x +8x; a=0
1 answer
1 answer