$$ 1 \leftarrow $$ For $$ f(x)=3 x^{4}-8 x^{3}+2 $$ find the following. (A) $$ f^{\prime}(x) $$ (B) The slope of the graph of $$ f $$ at $$ x=1 $$ (C) The equation of the tangent line at $$ x=1 $$ (D) The value(s) of $$ x $$ where the tangent line is horizontal

Question

UpStudy AI · Accepted Answer

**(A) Find $$ f'(x) $$.**

We have
$$
f(x)=3x^4-8x^3+2.
$$
Differentiate term-by-term:
$$
f'(x)=12x^3-24x^2.
$$

---

**(B) Find the slope of the graph at $$ x=1 $$.**

Evaluate $$ f'(1) $$:
$$
f'(1)=12(1)^3-24(1)^2=12-24=-12.
$$
Thus, the slope at $$ x=1 $$ is $$-12$$.

---

**(C) Find the equation of the tangent line at $$ x=1 $$.**

First, compute the point on the graph at $$ x=1 $$:
$$
f(1)=3(1)^4-8(1)^3+2=3-8+2=-3.
$$
The point is $$(1,-3)$$ and the slope is $$-12$$. Using the point-slope form of a line:
$$
y - (-3) = -12(x-1),
$$
which simplifies to
$$
y+3=-12(x-1).
$$
Solving for $$ y $$:
$$
y=-12x+12-3=-12x+9.
$$

---

**(D) Find the value(s) of $$ x $$ where the tangent line is horizontal.**

The tangent is horizontal when the slope is zero, so set $$ f'(x) $$ equal to zero:
$$
12x^3-24x^2=0.
$$
Factor out $$ 12x^2 $$:
$$
12x^2(x-2)=0.
$$
Set each factor equal to zero:
$$
12x^2=0 \quad \text{or} \quad x-2=0.
$$
This gives:
$$
x=0 \quad \text{or} \quad x=2.
$$
**(A) $$ f'(x) = 12x^3 - 24x^2 $$** **(B) The slope at $$ x=1 $$ is $$-12$$.** **(C) The equation of the tangent line at $$ x=1 $$ is $$ y = -12x + 9 $$.** **(D) The tangent line is horizontal at $$ x=0 $$ and $$ x=2 $$.**

For find the following.
(A)
(B) The slope of the graph of at
© The equation of the tangent line at
(D) The value(s) of where the tangent line is horizontal

Upstudy AI Solution

Answer

Solution

Mind Expander

Related Questions

Latest Calculus Questions

For find the following. (A) (B) The slope of the graph of at © The equation of the tangent line at (D) The value(s) of where the tangent line is horizontal