find a function f such that f(x) = 5x³ and the line 135x + y = 0 is tangent to the graph of f. f(x) =

Question

Accepted Answer

# Explanation:
## Step1: Integrate $ f^{\prime}(x) $
Integrate $ f^{\prime}(x)=5x^{3} $ using the power rule $ \int x^{n}dx=\frac{x^{n + 1}}{n+1}+C$ ($n
eq - 1$).
$$
\begin{align*}
f(x)&=\int5x^{3}dx\
&=5\times\frac{x^{3 + 1}}{3+1}+C\
&=\frac{5}{4}x^{4}+C
\end{align*}
$$
## Step2: Find the slope of the tangent line
Rewrite the line $135x + y=0$ in slope - intercept form $y=-135x$. The slope of this line is $m=-135$.
Since the slope of the tangent line to $y = f(x)$ is $f^{\prime}(x)$, set $f^{\prime}(x)=-135$.
$$
\begin{align*}
5x^{3}&=-135\
x^{3}&=-27\
x&=- 3
\end{align*}
$$
## Step3: Find the $y$ - coordinate of the tangent point
Substitute $x = - 3$ into the line equation $y=-135x$. So $y=-135\times(-3)=405$.
## Step4: Determine the value of $C$
Substitute $x=-3$ and $y = 405$ into $f(x)=\frac{5}{4}x^{4}+C$.
$$
\begin{align*}
405&=\frac{5}{4}\times(-3)^{4}+C\
405&=\frac{5}{4}\times81+C\
405&=\frac{405}{4}+C\
C&=405-\frac{405}{4}\
C&=\frac{1620 - 405}{4}\
C&=\frac{1215}{4}
\end{align*}
$$

# Answer:
$f(x)=\frac{5}{4}x^{4}+\frac{1215}{4}$

Sovi.AI - AI Math Tutor

QUESTION IMAGE

Question

Explanation:

Step1: Integrate \( f^{\prime}(x) \)

Step2: Find the slope of the tangent line

Step3: Find the \(y\) - coordinate of the tangent point

Step4: Determine the value of \(C\)

Answer: