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Answer from Sia

Posted 4 months ago

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.

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