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-7*x+(56/5)\t$-7x+\frac{56}{5}$
the answer above is not correct.
let $f(x)=\frac{x^{5}}{5}-4x^{2}+11$.
find an equation of the line tangent to the graph of $f$ at $x = 1$.
tangent line: $y=$

Explanation:

Step1: Find the derivative of $f(x)$

Using the power - rule $\frac{d}{dx}(x^n)=nx^{n - 1}$, if $f(x)=\frac{x^{5}}{5}-4x^{2}+11$, then $f^\prime(x)=\frac{5x^{4}}{5}-8x=x^{4}-8x$.

Step2: Evaluate the derivative at $x = 1$

Substitute $x = 1$ into $f^\prime(x)$: $f^\prime(1)=1^{4}-8\times1=1 - 8=-7$. This is the slope $m$ of the tangent line.

Step3: Find the $y$ - coordinate of the point on the function at $x = 1$

Substitute $x = 1$ into $f(x)$: $f(1)=\frac{1^{5}}{5}-4\times1^{2}+11=\frac{1}{5}-4 + 11=\frac{1}{5}+7=\frac{1 + 35}{5}=\frac{36}{5}$.

Step4: Use the point - slope form $y - y_1=m(x - x_1)$

Here $x_1 = 1$, $y_1=\frac{36}{5}$, and $m=-7$. So $y-\frac{36}{5}=-7(x - 1)$.
Expand the right - hand side: $y-\frac{36}{5}=-7x + 7$.
Add $\frac{36}{5}$ to both sides: $y=-7x+7+\frac{36}{5}=-7x+\frac{35 + 36}{5}=-7x+\frac{71}{5}$.

Answer:

$y=-7x+\frac{71}{5}$