Questions: Pergunta 4 A equação da reta tangente à curva f(x)=x^2 no ponto de abscissa igual a 3 é: (A) y=x (B) y=20x+6 (C) y=20-x (D) y=6x-9 (E) y=6x

Pergunta 4 A equação da reta tangente à curva f(x)=x^2 no ponto de abscissa igual a 3 é: (A) y=x (B) y=20x+6 (C) y=20-x (D) y=6x-9 (E) y=6x
Transcript text: Pergunta 4 A equação da reta tangente à curva $f(x)=x^{2}$ no ponto de abscissa igual a 3 é: (A) $y=x$ (B) $y=20 x+6$ (C) $y=20-x$ (D) $y=6 x-9$ (E) $y=6 x$
failed

Solution

failed
failed

Solution Steps

To find the equation of the tangent line to the curve f(x)=x2 f(x) = x^2 at the point where the x-coordinate is 3, we need to follow these steps:

  1. Calculate the derivative of f(x) f(x) to find the slope of the tangent line.
  2. Evaluate the derivative at x=3 x = 3 to get the slope at that point.
  3. Use the point-slope form of the equation of a line with the point (3,f(3)) (3, f(3)) and the slope found in step 2.
Step 1: Calcular a derivada da função f(x)=x2 f(x) = x^2

A derivada de f(x)=x2 f(x) = x^2 é: f′(x)=2x f'(x) = 2x

Step 2: Avaliar a derivada no ponto x=3 x = 3

Substituindo x=3 x = 3 na derivada: f′(3)=2⋅3=6 f'(3) = 2 \cdot 3 = 6 Portanto, a inclinação da reta tangente é 6 6 .

Step 3: Calcular a coordenada y y do ponto na curva

Substituindo x=3 x = 3 na função original: f(3)=32=9 f(3) = 3^2 = 9 Portanto, o ponto de tangência é (3,9) (3, 9) .

Step 4: Usar a forma ponto-inclinação para encontrar a equação da reta tangente

A forma ponto-inclinação da equação de uma reta é: y−y1=m(x−x1) y - y_1 = m(x - x_1) Substituindo m=6 m = 6 , x1=3 x_1 = 3 e y1=9 y_1 = 9 : y−9=6(x−3) y - 9 = 6(x - 3)

Step 5: Simplificar a equação da reta tangente

Simplificando a equação: y−9=6x−18 y - 9 = 6x - 18 y=6x−9 y = 6x - 9

Final Answer

y=6x−9\boxed{y = 6x - 9}

Was this solution helpful?
failed
Unhelpful
failed
Helpful

Questions asked by other users

failedAnswered
In Exercises 15-20, explain why S is not a basis for R^3. 17. S=(7,0,3),(8,-4,1)
failedAnswered
An analytical chemist weighs out 0.051 g of an unknown monoprotic acid into a 250 mL volumetric flask and dilutes to the mark with distilled water. He then titrates this solution with 0.0600 M NaOH solution. When the titration reaches the equivalence point, the chemist finds he has added 14.2 mL of NaOH solution. Calculate the molar mass of the unknown acid. Round your answer to 2 significant digits. g/mol
failedAnswered
Which species in the table below is the strongest reducing agent? Half-reaction Sn(aq)^(4+) + 2 e^- -> Sn(aq)^(2+) V^(3+)(aq) + e^- -> V(aq)^(2+) Zn(aq)^(2+) + 2 e^- -> Zn(s) I2(s) + 2 e^- -> 2 I(aq)
failedAnswered
Iron(III) sulfate is a(an) Ionic compound Molecular compound
failedAnswered
Given the following velocity vs. time graph, describe (in words) how you would move to match the motion described by the graph. Use words like faster, slower, toward, away, etc. Assume that a positive velocity is "forward" and a negative velocity is "backward".