Question

find an equation of the tangent line to the graph of the function f(x)=2e^{2x^{2}+3x} at x = 0. answer: y=

Explanation:

Step1: Find the function value at x = 0

Substitute \(x = 0\) into \(f(x)=2e^{2x^{2}+3x}\).

$$f(0)=2e^{2\times0^{2}+3\times0}=2e^{0}=2\times1 = 2$$

Step2: Differentiate the function using the chain - rule

Let \(u = 2x^{2}+3x\), then \(y = 2e^{u}\). The derivative of \(y\) with respect to \(u\) is \(\frac{dy}{du}=2e^{u}\), and the derivative of \(u\) with respect to \(x\) is \(\frac{du}{dx}=4x + 3\). By the chain - rule \(\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}=2e^{u}(4x + 3)=2e^{2x^{2}+3x}(4x + 3)\).

Step3: Find the slope of the tangent line at x = 0

Substitute \(x = 0\) into the derivative.

$$f^\prime(0)=2e^{2\times0^{2}+3\times0}(4\times0 + 3)=2\times1\times3=6$$

Step4: Use the point - slope form of a line

The point - slope form of a line is \(y - y_{1}=m(x - x_{1})\), where \((x_{1},y_{1})=(0,2)\) (from Step 1) and \(m = 6\) (from Step 3).

$$y-2=6(x - 0)$$

$$y=6x + 2$$

Answer:

\(y = 6x+2\)