Question

Find the tangent to the curve y=3x²+x+4 at x=3.

• A. 19
• B. 1.9
• C. 18
• D. 16

Answer 1

Answer: (A)

19

Answer 2

To find the tangent to the curve $y=3x^2+x+4$ at $x=3$, we need to find the slope of the tangent at that point.

1. Find the derivative of the curve: The derivative $\frac{dy}{dx}$ gives the slope of the tangent to the curve at any point $x$. Given the curve $y = 3x^2 + x + 4$. Differentiating with respect to $x$: $\frac{dy}{dx} = \frac{d}{dx}(3x^2) + \frac{d}{dx}(x) + \frac{d}{dx}(4)$ $\frac{dy}{dx} = 3(2x) + 1 + 0$ $\frac{dy}{dx} = 6x + 1$

2. Calculate the slope of the tangent at $x=3$: Substitute $x=3$ into the derivative expression to find the slope ($m$) of the tangent at that specific point. $m = \left.\frac{dy}{dx}\right|_{x=3} = 6(3) + 1$ $m = 18 + 1$ $m = 19$

   Thus, the slope of the tangent to the curve at $x=3$ is 19.

The slope of the tangent to a curve $y=f(x)$ at a point $x=a$ is given by the derivative $f'(a)$. For the given curve $y=3x^2+x+4$, the derivative is $\frac{dy}{dx} = 6x+1$. At $x=3$, the slope of the tangent is $6(3)+1 = 18+1 = 19$.