Chapter 4.4, Problem 29E

To determine

To calculate:The point at which the tangent to the curve is parallel to given line.

Answer to Problem 29E

The point is $x=\frac{\ln \left(5\right)-\ln \left(\ln \left(3\right)\right)}{\ln \left(3\right)},y=\frac{5}{\ln 3}+1$ .

Explanation of Solution

Given information:The equation of curve is,

  $y=3^x+1$

The equation of line is,

  $y=5x-1$

The slope of two parallel lines is same.

Formula used: $\left(slope\right):m_a=\frac{d{y}_a}{dx}$

Calculation:Slope of curve,

  $\begin{array}{l}{m}_1=\frac{dy}{dx}\\ m_1=\frac{d\left(3^x+1\right)}{dx}\\ m_1=3^x\times \ln \left(3\right)\\ \text{Calculate}\text{slope}\text{of}\text{line,}\\ m_2=\frac{dy}{dx}\\ m_2=\frac{d\left(5x-1\right)}{dx}\\ m_2=5\\ \text{Equate}\text{both}\text{slopes,}\\ m_1=m_2\\ 3^x\times \ln \left(3\right)=5\\ x=\frac{\ln \left(5\right)-\ln \left(\ln \left(3\right)\right)}{\ln \left(3\right)}\\ \text{Calculate}y,\\ y=\frac{5}{\ln 3}+1\end{array}$

The point at which the tangent to the curve is parallel to given line is $x=\frac{\ln \left(5\right)-\ln \left(\ln \left(3\right)\right)}{\ln \left(3\right)},y=\frac{5}{\ln 3}+1$ .