Answer to Question #219722 in Calculus for Heaven

Question

Find the critical numbers of the function

ƒ(x) = 2x³ +8x +x²

Find the absolute maximum and minimum values of f on the given interval

ƒ(x) = x^4 + 4x³ - 9

Verify the MVT and find the value of c in the given interval

 y = x³ - 9x² + 24x − 18; [2, 4]

Answer 1

1.

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

The critical numbers of the function are such numbers that $f'(x)=0.$

Find $f'(x)$

$$f'(x)=6x^2+2x+8$$

Solve for x

$$6x^2+2x+8=0\newline 3x^2+x+4=0$$

$D=1-4•3•4=-47, D<0,$

there are not real roots.

Answer. There aren't critical numbers.

2.

$$f(x)=x^4+4x^3-9$$

Find

$$f'(x)=4x^3+12x^2$$

Solve for x

$$4x^3+12x^2=0$$ $$4x^2(x+3)=0$$ $$x=0, \text{or } x=-3$$

There are two critical numbers 0 and -3.

Find the sign of derivative on the intervals $(-\infty,-3),(-3,0),(0,\infty).$

On the interval $(-\infty,-3)$ $f'(x)<0, f(x)$ decreases.

On the interval $(-3,0)$ $f'(x)>0, f(x)$ increases.

On the interval $(0,\infty)$ $f'(x)>0, f(x)$ increases.

Consider the segment [a,b], where a<-3 and b>0. Then the critical point -3 and 0 are in segment [a,b].

Let be a=-5 and b=1

Find

$f(-5)=(-5)^4+4\cdot(-5)^3-9=491 \newline f(0)=-9 \newline f(1)=1^4+4\cdot1^3-9=-4$

$f(-3)=(-3)^4+4•(-3)^3-9=-36.$

Answer. Therefore function $f(x)$ has the absolute minimum value -36 and the absolute maximum value 491 on the segment [-5,1].

3.

The function f is continuous on the interval [2,4] and differentiable on the interval ]2,4[, therefore we can apply the MVT (mean value theorem) which states that there is c∈[2,4]

 such that

$$f'(c)=\frac{f(4)-f(2)}{4-2}$$ $$f(x)=x^3-9x^2+24x-18$$ $$f(2)=2, f(4)=-2$$

Therefore,

$$f'(c)=(-2-2)/2=-2$$

Also,

$$f'(x)=3x^2-18x+24$$ $$f'(c)=3c^2-18c+24=-2$$

Solving for c

$$3c^2-18c+26=0$$

$D=12, D>0,$ so there are 2 real roots

$$c=\frac{9\pm\sqrt{3}}{3}.$$

Both values of c ∈[2,4].

Answer. $c=\frac{9\pm\sqrt{3}}{3}$