Question

find f(x) and find the equation of the line tangent to the graph of f at the indicated value of x. find the value(s) of x where the tangent line is horizontal. f(x)=3e^{x^{2}-7x + 5}, x = 0 f(x)=□

Explanation:

Step1: Apply chain - rule for differentiation

If \(y = 3e^{x^{2}-7x + 5}\), let \(u=x^{2}-7x + 5\), then \(y = 3e^{u}\). The derivative of \(y\) with respect to \(u\) is \(\frac{dy}{du}=3e^{u}\), and the derivative of \(u\) with respect to \(x\) is \(\frac{du}{dx}=2x - 7\). By the chain - rule \(\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}\), so \(f^{\prime}(x)=3e^{x^{2}-7x + 5}(2x - 7)\).

Step2: Find the slope of the tangent line at \(x = 0\)

Substitute \(x = 0\) into \(f^{\prime}(x)\): \(f^{\prime}(0)=3e^{0^{2}-7\times0 + 5}(2\times0 - 7)=3e^{5}\times(-7)=- 21e^{5}\).
Also, find \(f(0)\): \(f(0)=3e^{0^{2}-7\times0 + 5}=3e^{5}\).
The equation of the tangent line using the point - slope form \(y - y_{1}=m(x - x_{1})\) (where \((x_{1},y_{1})=(0,3e^{5})\) and \(m=-21e^{5}\)) is \(y-3e^{5}=-21e^{5}(x - 0)\), which simplifies to \(y=-21e^{5}x + 3e^{5}\).

Step3: Find when the tangent line is horizontal

Set \(f^{\prime}(x)=0\). Since \(e^{x^{2}-7x + 5}\gt0\) for all real \(x\), we set \(2x - 7 = 0\). Solving \(2x-7 = 0\) gives \(x=\frac{7}{2}\).

Answer:

\(f^{\prime}(x)=3(2x - 7)e^{x^{2}-7x + 5}\); Tangent line: \(y=-21e^{5}x + 3e^{5}\); Value of \(x\) for horizontal tangent: \(x = \frac{7}{2}\)