y=5e^kx

Question

y=5e^kx

Accepted Answer

∻Solution by Steps∻ ‖ step 1 ⋮ To find the value of $ k $, we need to determine the derivative of the given function $ y = (2x - 1)e^{kx} $ at $ x = 1 $. ‖ ‖ step 2 ⋮ The derivative of $ y $ with respect to $ x $ is $ y' = \frac{d}{dx}[(2x - 1)e^{kx}] $. Using the product rule, $ y' = (2)e^{kx} + (2x - 1)ke^{kx} $. ‖ ‖ step 3 ⋮ Evaluate the derivative at $ x = 1 $ to find the slope of the tangent line: $ y'(1) = (2)e^{k(1)} + (2(1) - 1)ke^{k(1)} = 2e^k + ke^k $. ‖ ‖ step 4 ⋮ The slope of the tangent line at $ x = 1 $ must be equal to the slope of the line $ y = 5e^x $ at $ x = 1 $, which is $ 5e $. ‖ ‖ step 5 ⋮ Set the expression for $ y'(1) $ equal to $ 5e $ and solve for $ k $: $ 2e^k + ke^k = 5e $. ‖ ‖ step 6 ⋮ Factor out $ e^k $ from the left side of the equation: $ e^k(2 + k) = 5e $. ‖ ‖ step 7 ⋮ Divide both sides by $ e $ to isolate $ e^{k-1}(2 + k) $: $ e^{k-1}(2 + k) = 5 $. ‖ ‖ step 8 ⋮ Since $ e^{k-1} = e^k/e $, we have $ (e^k/e)(2 + k) = 5 $. ‖ ‖ step 9 ⋮ Multiply both sides by $ e $ to get $ e^k(2 + k) = 5e $. ‖ ‖ step 10 ⋮ We already have $ e^k(2 + k) = 5e $ from step 6, so we can see that $ k $ must be such that $ 2 + k = 5 $. ‖ ‖ step 11 ⋮ Solve for $ k $: $ k = 5 - 2 $. ‖ ‖ step 12 ⋮ Simplify to find $ k $: $ k = 3 $. ‖ ∻Answer∻ ⚹ $ k = 3 $ ⚹ ∻Key Concept∻ ⚹Finding the slope of a tangent line to a curve at a given point⚹ ∻Explanation∻ ⚹The slope of the tangent line to the curve $ y = (2x - 1)e^{kx} $ at $ x = 1 $ is found by differentiating the function and evaluating it at $ x = 1 $. This slope must match the slope of the line $ y = 5e^x $ at $ x = 1 $, which leads to the equation $ e^k(2 + k) = 5e $. Solving this equation for $ k $ gives the value of $ k $ as 3.⚹◊What is the equation for finding the tangent line of a curve at a given point?⍭ Generate me a similar question◊

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