Questions: Let f(x)=4x+6-5e^x. Then the equation of the tangent line to the graph of f(x) at the point (0,1) is given by y=mx+b for m= b=

Let f(x)=4x+6-5e^x. Then the equation of the tangent line to the graph of f(x) at the point (0,1) is given by y=mx+b for m= b=
Transcript text: Let $f(x)=4 x+6-5 e^{x}$. Then the equation of the tangent line to the graph of $f(x)$ at the point $(0,1)$ is given by $y=m \boldsymbol{x}+\boldsymbol{b}$ for \[ \boldsymbol{m}= \] \[ b= \]
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Solution

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Solution Steps

Step 1: Find the derivative of \(f(x)\)

The derivative of \(f(x) = 4x + 6 - 5e^{x}\) with respect to \(x\) is \(f'(x) = 4 - 5e^{x}\).

Step 2: Calculate the slope at \(x_0\)

Substituting \(x_0 = 0\) into \(f'(x)\), we find the slope \(m = 4 - 5e^{0}\) = -1.

Step 3 & 4: Find the equation of the tangent line and calculate \(b'\)

Using the point-slope form, \(y - 1 = -1(x - 0)\), and simplifying to the slope-intercept form, we find \(b' = 1 + 1 \times 0 = 1\).

Final Answer:

The equation of the tangent line is \(y = -1.0x + 1\).

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