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