Question

find the equation of the line tangent to the graph of f at the indicated value of x. f(x)=5e^x + 2x; x = 0 y=

Explanation:

Step1: Find the derivative of f(x)

The derivative of $y = 5e^{x}+2x$ is $f^\prime(x)=5e^{x}+2$ using the derivative rules $\frac{d}{dx}(e^{x}) = e^{x}$ and $\frac{d}{dx}(ax)=a$.

Step2: Evaluate the derivative at x = 0

Substitute $x = 0$ into $f^\prime(x)$. So $f^\prime(0)=5e^{0}+2$. Since $e^{0}=1$, then $f^\prime(0)=5\times1 + 2=7$. This is the slope of the tangent - line.

Step3: Find the y - coordinate of the point of tangency

Substitute $x = 0$ into $f(x)$. So $f(0)=5e^{0}+2\times0$. Since $e^{0}=1$, then $f(0)=5$. The point of tangency is $(0,5)$.

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 $(x_{1},y_{1})=(0,5)$ and $m = 7$. Substituting these values gives $y - 5=7(x - 0)$.

Step5: Simplify the equation

$y-5 = 7x$ can be rewritten as $y=7x + 5$.

Answer:

$y = 7x+5$