Question

a. Find an equation of the line tangent to the given curve at a. b. Use a graphing utility to graph the curve and the tangent line on the same set of axes. $$y=e^{x} ; a=\ln 3$$

Short answer

Using a graphing utility, you will see that the tangent line intersects the curve at the point (ln(3), 3) and has a slope of 3 at that point. The tangent line approximates the curve closely only in the vicinity of this point. As you move away from this point, the tangent line and the curve will diverge.

Step-by-step solution

Find the derivative of the given function

To find the tangent line's equation, we first need to determine its slope. The slope of a tangent line to a curve is equal to the derivative of the function at that point. Differentiate y = e^x with respect to x: $$\frac{dy}{dx} = \frac{d(e^x)}{dx} = e^x$$

Evaluate the derivative at a = ln(3)

Now we need to find the slope of the tangent line at the given point a = ln(3). Plug a into the derivative: $$m = e^{ln(3)}$$ By properties of logarithms, e^(ln(3)) simplifies to 3: $$m = 3$$ So, the slope of the tangent line is 3.

Find the point on the curve where a = ln(3)

To find the tangent line's equation, we also need a point on the curve where a = ln(3). Plug ln(3) into the original function: $$y = e^{ln(3)}$$ As we saw earlier, e^(ln(3)) simplifies to 3: $$y = 3$$ So the point on the curve where a = ln(3) is (ln(3), 3).

Write the equation for the tangent line using the point-slope form

Now that we have the slope (m = 3) and the point (ln(3), 3), we can write the equation for the tangent line using the point-slope form: $$(y - 3) = 3(x - ln(3))$$ You can distribute the 3 to simplify the equation: $$y - 3 = 3x - 3ln(3)$$ Move the constant term to the other side of the equation: $$y = 3x - 3ln(3) + 3$$ The equation for the tangent line is: $$y = 3x - 3ln(3) + 3$$ a. The equation of the tangent line is y = 3x - 3ln(3) + 3. b. Use a graphing utility to graph the curve y = e^x and the tangent line y = 3x - 3ln(3) + 3 on the same set of axes.