find an equation of the line tangent to the following curve at the point (a,f(a)).

Question

f(x)=19e^x +6x; a=0

y=

Answer 1

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.