Question

If f(x) = 8x2 − x3, find f'(2) and use it to find an equation of the tangent line to the curve y = 8x2 − x3 at the point (2, 24).

Answer 1

f(x) = 8x2 − x3
f'(x) = 16x - 3x²

f'(2) = 16(2) - 3(2)²
f'(2) = 20

Using point-slope form:

y - y1 = m(x - x1)
y - 24 = 20 ( x - 2)

Therefore, the equation would be:

y = 20x - 16

Answer 2

f'(2) = 20

Equation of the tangent line → y = 20x - 16

Further explanation

To solve this problem there are several basic principles in Derivatives that need to be recalled, namely:

$y = a ~ x^n \Rightarrow \frac{dy}{dx} = a ~ n ~ x^{n-1}$

$y = \sin x \Rightarrow \frac{dy}{dx} = \cos x$

$y = \cos x \Rightarrow \frac{dy}{dx} = - \sin x$

$y = u \times v \Rightarrow \frac{dy}{dx} = u' \times v + u \times v'$

$y = u \div v \Rightarrow \frac{dy}{dx} = \frac{u' \times v - u \times v'}{v^2}$

$y = u^n \Rightarrow \frac{dy}{dx} = n \times u^{n-1} \times u'$

$\texttt{where u and v are functions in variable x}$

$\texttt{and u' and v' are derivatives of u and v}$

Let us now tackle the problem !

Given:

$f(x) = 8x^2 - x^3$

$f'(x) = 8(2x^{2-1}) - 3x^{3-1}$

$f'(x) = 16x^1 - 3x^2$

$f'(x) = 16x - 3x^2$

$f'(2) = 16(2) - 3(2)^2$

$f'(2) = 32 - 12$

$f'(2) = \boxed{20}$

Equation of the tangent line to the curve

$y - y_1 = m ( x - x_1 )$

$y - 24 = f'(2) ( x - 2 )$

$y - 24 = 20 ( x - 2 )$

$y - 24 = 20x - 40$

$y = 20x - 40 + 24$

$\boxed {y = 20x - 16}$

Learn more

• Implicit Differentiation : brainly.com/question/4711711
• Logarithmic Differentiation : brainly.com/question/9226310
• Calculus Problem : brainly.com/question/11237537

Answer details

Grade: High School

Subject: Mathematics

Chapter: Differentiation

Keywords: Maximum , Minimum , Value , Function , Variable , Derivation , Differentiation