To find the equation of the line tangent to the curve at the point (0, f(0)), we need to find the derivative of f(x) and evaluate it at x = 0.
The derivative of f(x) = 19e^x + 6x is:
f'(x) = 19e^x + 6
Now, we can evaluate f'(x) at x = 0 to find the slope of the tangent line:
f'(0) = 19e^0 + 6 = 19 + 6 = 25
So, the slope of the tangent line is 25.
Now, we can use the point-slope form of a linear equation to find the equation of the line tangent to the curve:
y - y1 = m(x - x1)
where (x1, y1) is the point (0, f(0)) and m is the slope of the tangent line.
Plugging in the values, we have:
y - f(0) = 25(x - 0)
Since x1 = 0 and f(0) = 19e^0 + 6(0) = 19 + 0 = 19, we have:
y - 19 = 25x
Rearranging the equation, we get:
y = 25x + 19
So, the equation of the line tangent to the curve f(x) = 19e^x + 6x at the point (0, f(0)) is y = 25x + 19.
find an equation of the line tangent to the following curve at the point (a,f(a)).
f(x)=19e^x +6x; a=0
y=
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