What is the equation of the line tangent to  f(x)=3x^2 + e^(1-x) at  x=1?

Dec 20, 2015

$y = 5 x - 1$

Explanation:

Step 1: Differentiate the function
$f ' \left(x\right) = \frac{d}{\mathrm{dx}} \left(3 {x}^{2}\right) + \frac{d}{\mathrm{dx}} \left({e}^{1 - x}\right)$

$= 6 x - {e}^{1 - x}$

Step 2: Find the slope of tangent line at $x = 1$
$m = f ' \left(1\right) = 6 \left(1\right) - {e}^{1 - 1} = 5$

Step 3: Find $y$ coordinate of the function when $x = 1$

$f \left(1\right) = 3 {\left(1\right)}^{2} + {e}^{1 - 1} = 3 + 1 = 4$

So, original point of the graph is $\left(1 , 4\right)$

To find equation of the line we can us point slope formula

$y - {y}_{1} = m \left(x - {x}_{1}\right)$ , when ${x}_{1}$ & ${y}_{1}$ is the point on original equation

$m = 5$ $\text{ " " " " " } \left(1 , 4\right)$

$y - 4 = 5 \left(x - 1\right)$
$\implies y = 5 x - 5 + 4$

$\implies y = 5 x - 1$